Introduction

This is a “follow-up” to the previous article: “The math exams of my life”, as some readers were curious to see some examples of Math Olympiad exercises.

This a selection of cute, non-trivial algebra problems (with a hint of number theory) compiled from the Romanian Math Olympiad (regional phase or faza judeteana) for 8th, 9th, and 10th graders (13-15 years old).

The solutions are surprising and involve a good understanding of algebraic concepts, pattern spotting, or tricks that, in the long run, help students develop mathematical intuition.

Depending on your passion for mathematics (or competitive mathematics), the problems should pose enough difficulty to keep you entertained for a few hours. If you are stuck with one problem, try to read the hint instead of going straight to the answer.

In case you want to solve them by yourself, do a short recap on the following subjects:

The main topic of this problem set is: “Inequalities”.

There’s a follow-up article - here.

I have a few notebooks containing solutions for various Math problems I’ve solved over the years (recreational mathematics). If time allows, I will publish more lists covering more topics. Currently, I am in the process of grouping them into categories.

“Screenshot” from one my notebooks.

The problems are from the “Regional” Phase of the Romanian Math Olympiad. The truly difficult problems are usually the ones from the “National” phase. I am planning to publish a list of those as well.

Another important aspect is that I am not a mathematician, so if you see that solutions are incorrect or have better solutions, please send me some feedback.

The Problems

1. easy Compute $$S=1-4+9-16+..+99^2-100^2$$. Hint & Answer

2. easy Determine the smallest element of the set $$\{ab \text{ } \lvert \text{ } a,b \in \mathbb{R} \text{ and } a^2 + 2b^2=1\}$$? Hint & Answer

3. easy What is the cardinality of following set: $$\{x \in \mathbb{R} \lvert [\frac{x+1}{5}]=\{\frac{x-1}{2}\} \}$$. $$\{ a \}$$ is the fractional part of the number $$a \in \mathbb{R}$$, and $$[a]$$ is the integer part of $$a \in \mathbb{R}$$. Hint & Answer

4. easy Find all the elements of the set $$\{\frac{3}{2x} \lvert x \in \mathbb{R} \text{ and, } \frac{1}{[x]} + \frac{1}{\{x\}}=2x \}$$. $$\{ a \}$$ is the fractional part of the number $$a \in \mathbb{R}$$, and $$[a]$$ is the integer part of $$a \in \mathbb{R}$$. Hint & Answer

5. easy Given $$a,b,c \in \mathbb{R}^{*}$$, we know that $$(a,b,c)$$ are part of an arithmetic progression, $$(ab, bc, ca)$$ are part a geometric progression, and $$a+b+c=ab+bc+ca$$. The triplets in the set $$M=\{(a,b,c)\ \lvert a,b,c \in \mathbb{R}^{*}\}$$ satisfy all the conditions mentioned before.

Compute $$\sum_{(a,b,c) \in M} (\lvert a \rvert + \lvert b \rvert + \lvert c \rvert)$$. Hint & Answer

6. easy For the following sequence of numbers $$(a_n)_{n \ge 1}$$, $$a_1=1$$ and $$a_2=6$$ and $$a_{n+1}=\frac{a_n}{a_{n-1}}$$, $$n \ge 2$$, compute $$a_{2021}$$. Hint & Answer

7. medium Find all numbers $$k \in \mathbb{Z}$$, so that $$a^4+b^4+c^4+d^4+k*abcd \ge 0$$, $$\forall a,b,c,d \in \mathbb{R}^{*}$$. Hint & Answer

8. medium If $$x,y,z,t \in \mathbb{R}$$, and $$(x-3y+6z-t)^2 \ge 2021$$ and $$x^2+y^2+z^2+t^2 \le 43$$, then what is the value of the expression: $$\lvert x + y + z + t \rvert$$ ? Hint & Answer

9. medium Considering $$x^2 + (a+b+c)x + k(ab+bc+ca) = 0$$ where $$a,b,c \in \mathbb{R}_{+}^{*} \text{, and } k \in \mathbb{R}$$ prove that $$\forall k \le \frac{3}{4}$$ the equation has all its solution in $$\mathbb{R}$$. Hint & Answer

10. easy Prove that $$[\frac{x+3}{6}] - [\frac{x+4}{6}] + [\frac{x+5}{6}] = [\frac{x+1}{2}] - [\frac{x+1}{3}]$$ is true, $$\forall x \in \mathbb{R}$$. Hint & Answer

11. medium Prove that if $$\sum_{k=1}^{n} a_k = \sum_{k=1}^{n} a_k^2$$ then $$\sum_{k=1}^{n} a_k \le n$$, with $$a_k \in \mathbb{R}_{+}$$. Hint & Answer

12. medium If $$a^2+b^2+c^2=3$$ prove ($$\lvert a \rvert + \lvert b \rvert + \lvert c \rvert - abc) \le 4$$, where $$a,b,c \in \mathbb{R}$$. Hint & Answer

13. hard Prove $$\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2} \ge 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$$ if $$a,b,c \in \mathbb{R}_{+}^{*}$$. Hint & Answer

14. easy Prove that if $$x \in \mathbb{R}$$ and $$x^2+x \in \mathbb{Q}$$ and $$x^3+2x \in \mathbb{Q}$$, then $$x \in \mathbb{Q} \subset \mathbb{R}$$. Hint & Answer

15. medium Given $$a,b \in \mathbb{R}$$, we know $$3^a+13^b=17^a$$, and $$5^a+7^b=11^b$$. Prove $$a \lt b$$. Hint & Answer

16. medium For $$n \in \mathbb{N}, n \ge 2$$, let $$u(n)$$ be the biggest prime number $$\le n$$ and $$v(n)$$ be the smallest prime number $$\gt n$$. Prove:

$\frac{1}{u(2)*v(2)}+\frac{1}{u(3)*v(3)}+...+\frac{1}{u(2010)*v(2010)}=\frac{1}{2}-\frac{1}{2021}$

17. medium Prove $$\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy} \le \frac{1}{2} (\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz})$$, $$\forall x,y,z \in \mathbb{R}_{+}^*$$. Hint & Answer

18. medium For $$a,b,c \in (0,1) \subset \mathbb{R}$$, $$x,y,z \in (0, \infty) \subset \mathbb{R}$$, if:

$\begin{cases} a^x = bc \\ b^y = ca \\ c^z = ab \end{cases}$

Prove that:

$\frac{1}{2+x} + \frac{1}{2+y} + \frac{1}{2+z} \le \frac{3}{4}$

19. easy If $$x,y,z \in R_{+}^{*}$$, and $$xy=\frac{z-x+1}{y}=\frac{z+1}{2}$$, prove that one of the numbers is the arithmetical mean of the other two. Answer

20. medium If $$a,b,c \in (1, \infty) or a,b,c \in (0,1)$$. Prove:

$log_a(bc) + log_b(ca) + log_c(ab) \ge 4(log_{ab}(c) + log_{bc}(a) + log_{ca}b)$

Hints

Hint 1.

Try playing with Faulhaber’s formula.

Hint 2.

Find a way to introduce $$ab$$ in the given equality $$a^2 + 2b^2=1$$.

Hint 3.

Try to get rid of the fractional part.

Hint 4.

Try to do some substitutions based on the fact $$x=[x]+\{x\}$$.

Hint 5.

Use $$a$$ to express $$b$$ and $$c$$.

Hint 6.

Look for any patterns by computing the first few terms of the sequence.

Hint 7.

Give meaningful values to $$a,b,c,d$$ and see what’s happening.

Have you considered AM-GM inequality?

Hint 8.

Can you use the Cauchy–Bunyakovsky–Schwarz inequality to solve the problem?

Hint 9.

Can you use the Rearrangement inequality to solve the problem?

Hint 10.

Can you use Hermite’s identity to solve the problem?

Hint 11.

Can you use the Cauchy–Bunyakovsky–Schwarz inequality to solve the problem?

Hint 12.

Can you use both CBS and AM-GM inequalities?

Hint 13.

Can you prove first $$\frac{x}{y^2}+\frac{y}{x^2} \ge \frac{1}{x} + \frac{1}{y}$$ ?

Hint 14.

Try expressing $$x$$ as a relationship between two rational numbers.

Hint 15.

Think in terms of monotonically increasing and monotonically decreasing functions.

Hint 16.

How many times a term $$\frac{1}{u(n)*v(n)}$$ appears in the sum ?

Hint 17.

Can you find a way to use the AM-GM inequality?

Hint 18.

Work on the expressions involving logarithms. Consider changing the base of the logarithms to a common one.

Hint 19.

This exercise is easy, so it doesn’t deserve a hint.

Hint 20.

Consider changing the base of the logarithms to a common one.

We write our sum as:

$S=(1^2-2^2)+(3^2-4^2)+...+(99^2-100^2)$

There is a formula for the difference of two square numbers:

$a^2-b^2=(a-b)(a+b)$

Our $$S$$ is a sum of differences between subsequent square numbers:

$S=(1-2)(1+2)+(3-4)(3+4)+...+(99-100)(99+100) \Leftrightarrow$ $S=-3-7-11-...-199$

If you have a keen eye for observations, you notice the numbers $$3,7,11,...,199$$ have this form $$3+k*4$$, $$k=0..49$$.

$S=-(3+0*4)-(3+1*4)-(3+2*4)-...-(3+49*4) \Leftrightarrow$ $S=-3*50-4(0+1+2+...+49) \Leftrightarrow$

The infinite series whose terms are natural numbers, $$1+2+3+...+n$$, is divergent. But we know the $$nth$$ partial sum of the series to be: $$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}$$, so:

$S=-150-4*\frac{49(49+1)}{2}$

The final answer $$S=-150-4900=-5050$$.

If you are familiar with Faulhaber’s formula, you know that:

$\sum_{k=1}^n k = \frac{n(n+1)}{2}$ $\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$

We can cleverly use the two formulas and reimagine our $$S$$ to be:

$S=\underbrace{(1^2+2^2+3^2+...+100^2)}_{\sum_{k=1}^{100} k^2} - \underbrace{2*(2^2+4^2+6^2+..+100^2)}_{\sum_{k=1}^{50} (2k)^2} \Leftrightarrow$ $S=\frac{100*101*201}{6}-2*(2^2*1^2+2^2*2^2+2^2*3^2+...+2^2*50^2) \Leftrightarrow$ $S=\frac{100*101*201}{6}-8*\frac{50*51*101}{6}$

Final answer $$S=-5050$$.

The intuition begs us to find a way to link our existing relationship to a term containing $$ab$$. In this regard:

$1=a^2+2b^2=a^2 + (b\sqrt{2})^2 \Leftrightarrow$ $1=a^2 + \underbrace{2*\sqrt{2}ab}_{\text{we add this}} + (b\sqrt{2})^2 - \underbrace{2*\sqrt{2}ab}_{\text{to remove it later}} \Leftrightarrow$ $1=(a+b\sqrt{2})^2 - 2\sqrt{2}ab \Leftrightarrow$ $1+2\sqrt{2}ab=(a+b\sqrt{2})^2$

But we know that:

$(a+b\sqrt{2})^2 \geq 0 \Rightarrow$ $1+2\sqrt{2}ab \geq 0 \Rightarrow$ $ab \geq \frac{-1}{2\sqrt{2}}$

But is it possible for $$ab=\frac{-1}{2\sqrt{2}}$$. Yes, $$ab=\frac{-1}{2\sqrt{2}}$$ when $$(a+b\sqrt{2})^2=0$$.

All in all, the smallest element of our set is $$\frac{-1}{2\sqrt{2}}$$.

If $$a \in \mathbb{R}$$, then:

$a=[a]+\{a\} \Leftrightarrow$ $\{a\}=a-[a]$

With this in mind, we want to get rid of the wild fractional part $$\{a\}$$ from our relationship:

$[\frac{x+1}{5}]=\frac{x-1}{2} - [\frac{x-1}{2}] \Leftrightarrow$ $\underbrace{[\frac{x+1}{5}]}_{\in \mathbb{Z}}+\underbrace{[\frac{x-1}{2}]}_{\in \mathbb{Z}}=\frac{x-1}{2}$

We can safely say that $$\frac{x-1}{2}=k \in \mathbb{Z}$$.

We substitute $$x=2k+1$$ in the original relation:

$[\frac{2*k+1+1}{5}] + [\frac{2*k+1-1}{2}] = k \Leftrightarrow$ $[\frac{2k+2}{5}] + k = k \Leftrightarrow$ $[\frac{2k+2}{5}] = 0$

Because $$[\frac{2k+2}{5}] = 0$$, then:

$0 \leq \frac{2k+2}{5} \lt 1 \Leftrightarrow$ $-1 \leq k \lt \frac{3}{2}$

There are $$3$$ numbers $$k \in \mathbb{Z}$$ that satisfy the relationship: $$\{-1,0,1\}$$.

But, $$x=2k+1 \Rightarrow x \in \{-1, 1, 3\}$$.

Testing our solutions:

$x = -1 \Rightarrow [\frac{-1+1}{5}] = \{ \frac{-1-1}{2} \} \Leftrightarrow [0] = \{ -1 \} \text{ is true}$ $x = 1 \Rightarrow [\frac{1+1}{5}] = \{ \frac{1-1}{2} \} \Leftrightarrow [\frac{2}{5}] = \{ 0 \} \text{ is true}$ $x = 1 \Rightarrow [\frac{3+1}{5}] = \{ \frac{3-1}{2} \} \Leftrightarrow [\frac{4}{5}] = \{ 1 \} \text{ is true}$

So the cardinality is $$3$$.

To avoid division by zero in $$\frac{3}{2x}$$, we know that $$x \neq 0$$.

Note: The fractional part operation (denoted by $$\{a\}$$) is not distributive over multiplication, meaning $$\{a*b\} \neq \{a\} * \{b\}$$, so avoid making any further assumptions about $$\frac{1}{\{x\}}$$.

Firstly, $$\frac{1}{[x]} + \frac{1}{\{x\}}=2x$$ is equivalent to $$\frac{x}{[x]\{x\}}=2x$$, is equivalent to $$x(2[x]\{x\}-1)=0$$.

Because $$x \neq 0$$, then $$2[x]\{x\}=1$$, or $$\{x\}=\frac{1}{2[x]}$$.

We substitute $$[x]=n$$, so that $$\{x\}=\frac{1}{2n}$$, where $$n \in \mathbb{Z}$$ and $$n \geq 1$$.

$$x=[x]+{x}$$ becomes $$x=n+\frac{1}{2n}=\frac{2n^2+1}{2n}$$.

The term defining the set becomes $$[\frac{3}{2x}] \rightarrow [\frac{3}{2*\frac{2n^2+1}{2n}}]=[\frac{3n}{2n^2+1}]$$.

For $$n \ge 1$$, we know that $$0 \lt \frac{3n}{2n^2+1} \lt 1$$. Without calculus, we can easily prove that for $$n \ge 1$$, $$f(n)=3n$$, and $$g(n)=2n^2+1$$ are increasing functions, with $$g(n)$$ increasing faster, so that $$\frac{f(n)}{g(n)} \lt 1$$.

For certain we know that $$0 \lt \frac{3n}{2n^2+1}$$.

After we solve $$\frac{3n}{2n^2+1} \lt 1$$, we obtain $$n \ge 1$$.

For $$n=1$$, $$[\frac{3}{2x}] \rightarrow 1$$, for $$n \gt 1$$, $$[\frac{3}{2x}] \rightarrow 0$$.

So the set $$\{\frac{3}{2x} \lvert x \in \mathbb{R} \text{ and, } \frac{1}{[x]} + \frac{1}{\{x\}}=2x \} = \{0, 1\}$$

$$(a,b,c)$$ are in an arithmetic progression $$\Rightarrow b=\frac{a+c}{2}$$.

$$(ab, bc, ac)$$ are in a geometric progression $$\Rightarrow (bc)^2 = ab*ac$$.

With this in mind, we have the following relationships:

$\begin{cases} b=\frac{a+c}{2} \\ (bc)^2 = ab*ac \\ a+b+c=ab+bc+ca \end{cases}$

It doesn’t look like it, but we have enough information to determine $$(a,b,c)$$

$\begin{cases} 2b=a+c \\ bc=a^2 \Leftrightarrow c=\frac{a^2}{b}\\ 3b=ab+bc+ac \end{cases}$

But we can also write $$2b=a+\frac{a^2}{b}$$.

Putting all together in the last equation:

$3b=ab+\underbrace{bc}_{a^2}+a\underbrace{c}_{\frac{a^2}{b}} \Leftrightarrow \\ 3b=ab+a^2+\frac{a^3}{b} \Leftrightarrow \\ 3b=a(\underbrace{a+\frac{a^2}{b}}_{2b})+ab \Leftrightarrow \\ 3b=2ab+ab$

But because $$b \in \mathbb{R}^{*}$$, and $$3b=3ab$$, we can safely assume $$a=1$$.

If $$a=1$$, the relationship $$2b=a+c$$ becomes $$2b=1+\frac{1}{b} \Leftrightarrow 2b^2-b-1=0$$.

$$2b^2-b-1=0$$ can also be written as $$2b^2+b-(2b+1)=0$$ or $$b(2b+1)-(2b+1)=0 \Leftrightarrow (b-1)(2b+1)=0$$, so $$b=1$$ or $$b=-\frac{1}{2}$$.

For $$b=1$$, $$c=1$$, and for $$b=-\frac{1}{2}$$, $$c=-2$$.

So our triplets are $$(1,1,1)$$ or $$(1,-\frac{1}{2}, -2)$$.

Computing the sum is trivial: $$\sum=\frac{13}{2}$$.

The key to solving this exercise is to compute the first terms of the sequence:

$\begin{cases} a_1=2 \\ a_2=6 \\ a_3=\frac{6}{2}=3 \\ a_4=\frac{3}{6}=\frac{1}{2} \\ a_5=\frac{\frac{1}{2}}{3}=\frac{1}{6} \\ a_6=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{3} \\ a_7=\frac{\frac{1}{3}}{\frac{1}{6}}=2 \\ a_8=\frac{2}{\frac{1}{3}}=6 \\ \text{... and so on} \end{cases}$

We can see that after every 6 terms, the values repeat themselves. Using mathematical induction, we can prove that our sequence is:

$S(n)= \begin{cases} 2 \text{ ,}\text{ if }n=6*k+1 \\ 6 \text{ ,}\text{ if }n=6*k+2 \\ 3 \text{ ,}\text{ if }n=6*k+3 \\ \frac{1}{2} \text{ ,}\text{ if }n=6*k+4 \\ \frac{1}{6} \text{ ,}\text{ if }n=6*k+5 \\ \frac{1}{3} \text{ ,}\text{ if }n=6*k \end{cases} \text{, where } k \in \mathbb{N}$

Final answer is: $$a_{2021} = \frac{1}{6}$$

When you have problems like this, you need to consider giving specific values to $$a,b,c,d$$ that can help you narrow down your search.

For example, we know the inequality should hold regardless of $$a, b, c, d$$ ($$\forall$$), so let’s assume the following:

$a=b=c=d=n$

Then:

$4*n^4+k*n^4 \ge 0 \Leftrightarrow \\ k \ge -4$

This already tells us that we need to look somewhere in $$k \in [-4, \infty) \cap \mathbb{Z}$$.

On my first try, I’ve tried with $$a^2=b^2=m$$ and $$c^2=b^2=n$$, so I’ve got something like $$(n\sqrt{2} - m\sqrt{2})^2 + mn*(k+4) \ge 0$$.

Now, we need to find an upper bound for $$k$$. For this, we need to apply some tricks to obtain:

$\text{something positive} + (\text{an upper bound}-k)*(\text{something positive}) \ge 0$

So why don’t we pick:

$\begin{cases} a=b=c=n \\ d=-n \end{cases}$

Then our relationship becomes:

$4*n^4 - k*n^4 \ge 0 \Leftrightarrow \\ k \le 4$

So, at this point, we know that $$k \in [-4, 4] \cap \mathbb{Z}$$. Some would stop here, and that would be wrong.

We need to come up with stronger proof. Our findings, $$k \in [-4, 4] \cap \mathbb{Z}$$ were based on specific values for $$a,b,c,d$$.

One famous inequality in mathematics is the inequality of arithmetic and geometric means:

$(\frac{x_1^n+x_2^n+...+x_n^n}{n})^{\frac{1}{n}} \ge ... \ge \frac{x_1+x_2+...+x_n}{n} \ge (x_1 * x_2 * ... * x_n)^{\frac{1}{n}}$

If you are good at spotting patterns, you will see how it resembles our inequality.

$\begin{cases} x_1 \rightarrow \lvert a \rvert \\ x_2 \rightarrow \lvert b \rvert \\ x_3 \rightarrow \lvert c \rvert \\ x_4 \rightarrow \lvert d \rvert \end{cases}$

So, if, we consider this:

$(\frac{\lvert a \rvert ^4 + \lvert b \rvert ^4 +\lvert c \rvert ^4+\lvert d \rvert ^4}{4})^{\frac{1}{4}} \ge (\lvert a \rvert * \lvert b \rvert * \lvert c \rvert * \lvert d\rvert)^{\frac{1}{4}} \Leftrightarrow \\ a^4+b^4+c^4+d^4 \ge 4 * \lvert abcd \rvert$
• If $$abcd \ge 0$$
• Then $$a^4+b^4+c^4+d^4 + k*abcd \ge 4*abcd + k*abcd$$;
• But $$abcd*(k+4) \ge 0$$, for $$k \in [-4,4] \cap \mathbb{Z}$$;
• If $$abcd \lt 0$$
• Then $$a^4+b^4+c^4+d^4 + k*abcd \ge -4*abcd + k*abcd$$;
• But $$-abcd(4-k) \ge 0$$, for $$k \in [-4, 4] \cap \mathbb{Z}$$.

This tells us that the inequality works $$\forall k \in -4, 4 \cap \mathbb{Z}$$.

The first observation you should make is that $$2021=43*41$$.

When we have problems like this, it’s worth checking if using the two fundamental inequalities, AM-GM inequality and Cauchy–Bunyakovsky–Schwarz helps us.

Even if not fully obvious in our case, the one that helps is the CBS:

$(\sum_{i=1}^{n} a_i^2) * (\sum_{i=1}^{n}b_i^2) \ge (\sum_{i=1}^{n}a_i*b_i)^2 \\ \text{ equality holds if } \frac{a_i}{b_i}=k \text{, for } k \in \mathbb{R}$

For 3 pair of numbers, $$(a_1, b_1), (a_2, b_2), (a_3, b_3)$$, CBS looks like this:

$(a_1*b_1 + a_2*b_2 + a_3*b_3)^2 \le (a_1^2 + a_2^2 + a_3^2) * (b_1^2 + b_2^2 + b_3^2)$

Firstly, we know that: $$2021 \le (1*x+(-3)y+6z+(-1)*t)^2$$

So, thinking in terms of the CBS inequality, why don’t we consider the following:

$\begin{cases} a_1 \rightarrow x \\ a_2 \rightarrow y \\ a_3 \rightarrow z \\ a_4 \rightarrow t \\ \end{cases}$

and

$\begin{cases} b_1 \rightarrow 1 \\ b_2 \rightarrow -3 \\ b_3 \rightarrow 6 \\ b_4 \rightarrow -1 \\ \end{cases}$

In this regard, we can write things like this:

$2021 \le (x-3y+6z+t)^2 \le \underbrace{(1^2+(-3)^2+6^2+1^2)}_{47}*\underbrace{(x^2+y^2+z^2+t^2)}_{\le 43} \le 2021$

This means our expression, $$(x-3y+6z+t)^2$$ is squeezed between $$2021$$ and $$2021$$, so the equality holds true, so:

$\frac{x}{1}=\frac{y}{-3}=\frac{z}{6}=\frac{t}{-1}=k \text{ , } k \in \mathbb{R}$

Or:

$\begin{cases} x=k \\ y=-3k \\ z=6k \\ t=-k \end{cases} \text{ , } k \in \mathbb{R}$

If we use this substitution:

$x^2+y^2+z^2+t^2 \le 43 \Leftrightarrow \\ k^2 + 9k^2+36k^2+k^2 \le 43 \\ 47k^2 \le 43 \\ k \le \sqrt{\frac{43}{47}}$

And then again:

$(x-3y+6z-t)^2 \ge 2021 \\ (k+9k+36k+k)^2 \ge 43*47 \\ 47^2*k^2 \ge 43*47 \\ k^2 \ge \frac{43}{47} \\ k \ge \sqrt{\frac{43}{47}}$

So we can safely assume that our $$k=\sqrt{\frac{43}{47}}$$.

Now it’s easy to compute the expression: $$\lvert x^2 + y^2 + z^2 + t^2 \rvert=3 * \sqrt{\frac{43}{47}}$$.

For 6 numbers: $$x_1 \le x_2 \le x_3$$ and $$y_1 \le y_2 \le y_3$$ the Rearrangement inequality can be written as:

$x_1*y_2 + x_2*y_3+x_3*y_1 \le x_1*y_1 + x_2*y_2 + x_3*y_3$

If we pick:

$\begin{cases} x_1=y_1=a \\ x_2=y_2=b \\ x_3=y_3=c \\ \end{cases}$

We obtain:

$ab+bc+ca \le a^2 + b^2 + c^2 (*)$

Then:

$2(ab+bc+ca) + ab+bc+ca \le 2(ab+bc+ca) + a^2 + b^2 + c^2 \Leftrightarrow \\ 3(ab+bc+ca) \le (a+b+c)^2 (**)$

Now, let’s get to our problem. We will compute the $$\Delta$$ for $$x^2 + (a+b+c)x + k(ab+bc+ca) = 0$$:

$\Delta=(a+b+c)^2-4k(ab+bc+ca)$

For our equation to have solutions in $$\mathbb{R}$$, we need to find $$k$$ so that $$\Delta \ge 0$$.

$\Delta = \underbrace{(a+b+c)^2}_{\ge 3(ab+bc+ca)} - 4k(ab+bc+ca) \ge 0 \\$

Using (**), we can write:

$\Delta \ge 3(ab+bc+ca) - 4k(ab+bc+ca) \ge 0 \Leftrightarrow \\ (3-4k) \ge 0 \Rightarrow k \le \frac{3}{4}$

Hermite’s Identity states that:

$\sum_{k=0}^{n-1}[x+\frac{k}{n}] = [nx] \text{ , } \forall x \in \mathbb{R} \text{ ,and } n \in \mathbb{N}$

One idea is to make our existing terms (e.g. [$$\frac{x+3}{6}$$]) resemble the terms to Hermite’s identity ($$[x+\frac{k}{n}]$$). In this regard we need to find a way to “isolate” the $$x$$ outside the fraction(s).

One idea is to perform the following substitution:

$y=\frac{x+1}{6}$

So our identity becomes:

$[y+\frac{1}{3}] - [y+\frac{1}{2}] + [y+\frac{2}{3}] = [3y] - [2y] \text{ (*)}$

But:

$\begin{cases} [3y] = [y] + [y+\frac{1}{3}] + [y+\frac{2}{3}] \\ [2y] = [y] + [y+\frac{1}{2}] \end{cases} \text{ (**)}$

(*) and (**) proves the identity to be correct.

Even if it’s not obvious, let’s start again with the CBS inequality:

$(\sum_{i=1}^{n} a_i^2) * (\sum_{i=1}^{n}b_i^2) \ge (\sum_{i=1}^{n}a_i*b_i)^2 \\$

If we pick $$b_1=b_2=b_3=...=b_n=1$$ the inequality becomes:

$(\sum_{i=1}^{n} a_i^2) * n \ge (\sum_{i=1}^{n}a_i)^2$

Expanding the sums:

$(a_1+a_2+...+a_n)^2 \le n*(a_1^2 + a_2^2 + ...+ a_n^2)$

$$a_k \in \mathbb{R}_{+}$$ so we can conclude:

$a_1+a_2+...b_n \le n$

12. easy If $$a^2+b^2+c^2=3$$ prove ($$\lvert a \rvert + \lvert b \rvert + \lvert c \rvert - abc) \le 4$$, where $$a,b,c \in \mathbb{R}$$. Hint & Answer

The following is true ($$\forall a,b,c \in \mathbb{R}$$):

$a^2 + b^2 + c^2 = \lvert a \rvert^2 + \lvert b \rvert^2 + \lvert c \rvert^2 = 3$

Applying Cauchy–Bunyakovsky–Schwarz:

$\underbrace{(a^2 + b^2 + c^2)}_{=3} * \underbrace{(1^2 + 1^2 + 1^2)}_{=3} \ge (\lvert a \rvert*1 + \lvert b \rvert*1 + \lvert c \rvert*1)^2$

From this $$\Rightarrow 3 \ge (\lvert a \rvert + \lvert b \rvert + \lvert c \rvert)$$ (*).

Applying AM-GM inequality:

$\frac{a^2+b^2+c^2}{3} \ge (abc)^{\frac{2}{3}}$

We can safely observe $$-abc \le 1$$ (**).

(*) and (**) $$\Rightarrow$$:

$\underbrace{\lvert a \rvert + \lvert b \rvert + \lvert c \rvert}_{\le 3} \underbrace{- abc}_{\le 1} \le 4 \text{ is TRUE}$

Looking at:

$\frac{a+b}{c^2}+\frac{b+c}{a^2}+\frac{c+a}{b^2} \ge 2(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

We can group things in the following manner:

$\underbrace{\frac{a}{b^2}+\frac{b}{a^2}}_{*} + \underbrace{\frac{b}{c^2}+\frac{c}{b^2}}_{**} + \underbrace{\frac{a}{c^2} + \frac{c}{a^2}}_{***} \ge \underbrace{\frac{1}{a} + \frac{1}{b}}_{*} + \underbrace{\frac{1}{b} + \frac{1}{c}}_{**} + \underbrace{\frac{1}{a} + \frac{1}{c}}_{***}$

There’s a pattern here!! If we manage to solve this inequality: $$\frac{m}{n^2} + \frac{n}{m^2} \ge \frac{1}{n} + \frac{1}{m}$$, $$m,n \in \mathbb{R}_{+}^*$$ we will be able solve our problem.

So let’s solve this:

$\frac{m}{n^2} + \frac{n}{m^2} \ge \frac{1}{n} + \frac{1}{m} \Leftrightarrow \\$ $m^3 + n^3 = mn(m+n) \ge 0 \Leftrightarrow \\$ $m^2(m-n)-n^2(m-n) \ge 0 \Leftrightarrow \\$ $(m-n)(m^2-n^2) \ge 0 \Leftrightarrow \\$ $\underbrace{(m-n)^2}_{\ge 0} * \underbrace{(m+n)}_{\gt 0} \ge 0$

Now, we know the following:

$\begin{cases} \frac{a}{b^2}+\frac{b}{a^2} \ge \frac{1}{a} + \frac{1}{b} \\ \frac{b}{c^2}+\frac{c}{b^2} \ge \frac{1}{b} + \frac{1}{c} \\ \frac{a}{c^2}+\frac{c}{a^2} \ge \frac{1}{a} + \frac{1}{c} \end{cases}$

If we sum all three inequalities, we’ve proven the original inequality.

If $$x^2+x \in \mathbb{Q}$$ , then $$x^2+x=a \in \mathbb{Q}$$

If $$x^3+2x \in \mathbb{Q}$$ , then $$x^3+2x=b \in \mathbb{Q}$$.

Our purpose is to try expressing $$x$$ using only $$a$$ and $$b$$ in a way we will prove $$x \in \mathbb{Q}$$.

We start by doing some tricks with $$b=x^3+2x=x^3+\underbrace{(x^2-x^2)}_{=0}+\underbrace{(x-x)}_{=0}+2x$$.

After regrouping terms, $$b$$ becomes $$b=x\underbrace{(x^2+x)}_{=a}-\underbrace{(x^2+x)}_{=a} + x + 2x$$.

So $$b=x(a+3)-a$$, or $$a+b=x(a+3)$$.

We are getting closer to the solution; the only thing remaining is to check if $$a=-3$$.

Let’s suppose $$a=-3$$, then $$x^2+x+3=0$$, but this is impossible because the solutions $$x_1,x_2 \notin \mathbb{R}$$. So, we can safely assume $$a \neq 3$$.

If $$a \neq 3$$, then we can write $$x=\frac{a+b}{a+3}$$. But both $$a+b \in Q$$ and $$a+3 \in Q$$, then for sure $$x \in Q$$.

The fact that $$3,5,7,11,13$$ are all prime numbers is just a coincidence, but congratulations if you spot that.

Sometimes, it is easier to disprove, than to prove something, so ad absurdum let’s suppose $$a \ge b$$.

Firstly, for the function $$f(x)=a^x$$, if $$a$$ is a constant greater than 1, we say that the function is increasing, meaning the value of $$f(x)$$ increases with $$x$$. Secondly, if $$a \lt 1$$, then we say the function is decreasing, meaning the value of $$f(x)$$ decreases while $$x$$ increases.

That being said, if $$a \ge b$$, then $$13^a \ge 13^b$$ and $$5^a \ge 5^b$$.

This means that $$3^a + 13^a \ge 17^a$$, or $$(\frac{3}{17})^a + (\frac{13}{17})^a \ge 1$$.

Let $$g(x) : \mathbb{R} \rightarrow \mathbb{R}$$, $$g(x)=(\frac{3}{17})^x + (\frac{13}{17})^x$$. $$g(x)$$ is stricly decreasing. Also, $$g(1)=\frac{16}{17} \lt 1$$. But $$g(1) \lt g(a)$$ thus, $$a \lt 1$$ (*).

If our initial supposition is correct, then $$5^b+7^b \ge 11^b$$. Following the same principle, we define $$h(x)=(\frac{5}{11})^b+(\frac{7}{11})^b \ge 1$$ and we eventually conclude that $$b \gt 1$$ (**).

(*) and (**) $$\Rightarrow$$ our supposition is false, so $$a \lt b$$ is true.

This problem is easier than it looks.

$S=\underbrace{\frac{1}{u(2)*v(2)}}_{=\frac{1}{2*3}}+\underbrace{\frac{1}{u(3)*v(3)}}_{=\frac{1}{3*5}}+\underbrace{\frac{1}{u(4)*v(4)}}_{=\frac{1}{3*5}}+\underbrace{\frac{1}{u(5)*v(5)}}_{=\frac{1}{5*7}}+\underbrace{\frac{1}{u(6)*v(6)}}_{=\frac{1}{5*7}}+\underbrace{\frac{1}{u(7)*v(7)}}_{=\frac{1}{7*11}}...$

We see that the terms in our sum repeat themselves a number of times.

For example:

• $$\frac{1}{2*3}$$ appears once;
• $$\frac{1}{3*5}$$ appears two times;
• $$\frac{1}{5*7}$$ appears two times;
• $$\frac{1}{7*11}$$ will appear four times;
• … and so on.

If $$p, q$$ are consecutive prime numbers, we define the set $$M=\{ n \in \mathbb{N} \lvert q \le n \lt p \}$$. Th cardinal of $$M$$ is exactly $$p-q$$ for a given $$n$$. It means that each term $$\frac{1}{qp}$$ appears exactly $$p-q$$ times in our sum.

With this in mind, we can re-write our sum as:

$S=\frac{3-2}{2*3}+\frac{5-3}{5*3}+\frac{7-5}{5*7}+\frac{11-7}{7*11}+... \Leftrightarrow$ $S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2011} \Leftrightarrow$ $S=\frac{1}{2}-\frac{1}{2011}$

We use the AM-GM inequality for each of the following terms:

$x^2+yz \le 2\sqrt{x^2yz} = 2x\sqrt{yz} \Leftrightarrow \\$ $\frac{1}{x^2+yz} \le \frac{\sqrt{yz}}{2xyz} (*) \\$

But $$\sqrt{yz} \le \frac{y+z}{2} (**)$$.

$$(*), (**) \Rightarrow$$:

$\frac{1}{x^2+yz} \le \frac{\frac{y+z}{2}}{2xyz}$

Similarly, for the other terms:

$\frac{1}{y^2+xz} \le \frac{\frac{x+z}{2}}{2xyz}$ $\frac{1}{z^2+xy} \le \frac{\frac{x+y}{2}}{2xyz}$

If we sum everything up:

$\frac{1}{x^2+yz} + \frac{1}{y^2+xz} + \frac{1}{z^2+xy} \le \frac{\frac{y+z}{2}}{2xyz} + \frac{\frac{x+z}{2}}{2xyz} + \frac{\frac{x+y}{2}}{2xyz} \Leftrightarrow \\$ $\frac{1}{x^2+yz} + \frac{1}{y^2+xz} + \frac{1}{z^2+xy} \le \frac{1}{2}(\frac{x+y+z}{xyz}) \Leftrightarrow \\$ $\frac{1}{x^2+yz} + \frac{1}{y^2+xz} + \frac{1}{z^2+xy} \le \frac{1}{2} (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{xz})$

Let’s start by working on the expressions involving logarithms.

$\begin{cases} a^x = bc \Leftrightarrow log_a(a^x) = log_a(bc) \Leftrightarrow x = log_a(b) + log_a(c) \\ b^y = ca \Leftrightarrow log_b(b^y) = log_b(ca) \Leftrightarrow y = log_b(c) + log_b(a) \\ c^z = ab \Leftrightarrow log_c(c^z) = log_c(ab) \Leftrightarrow c = log_c(a) + log_c(b) \\ \end{cases}$

Now, let’s change the base for our logarithms to a common number, $$m \in \mathbb{R}_{+}^{*}$$:

$\begin{cases} x = log_a(b) + log_a(c) = \frac{log_m(b)}{log_m(a)} + \frac{log_m(c)}{log_m(a)} \\ y = log_b(c) + log_b(a) = \frac{log_m(c)}{log_m(b)} + \frac{log_m(a)}{log_m(b)} \\ c = log_c(a) + log_c(b) = \frac{log_m(a)}{log_m(c)} + \frac{log_m(b)}{log_m(c)} \\ \end{cases}$

Let’s define $$l_a = log_m(a)$$, $$l_b = log_m(b)$$, $$l_c = log_m(c)$$.

We observe that, $$x,y,z$$ can be written as:

$\begin{cases} x=\frac{l_b+l_c}{l_a} \\ y=\frac{l_c+l_a}{l_b} \\ z=\frac{l_a+l_b}{l_c} \end{cases}$

The expression required to be proven becomes:

$\frac{1}{2+\frac{l_b+l_c}{l_a}}+\frac{1}{2+\frac{l_c+l_a}{l_b}}+\frac{1}{2+\frac{l_a+l_b}{l_c}} \le \frac{3}{4} \Leftrightarrow$ $\frac{l_a}{l_a + \underbrace{(l_a + l_b + l_c)}_{s_l}} + \frac{l_b}{l_b+\underbrace{(l_a+l_b+l_c)}_{s_l}} + \frac{l_c}{l_c+\underbrace{(l_a+l_b+l_c)}_{s_l}} \le \frac{3}{4} \Leftrightarrow$ $1-\frac{l_a}{l_a+s_l}+1-\frac{l_b}{l_b+s_l}+1-\frac{l_c}{l_c+s_l} \ge 3-\frac{3}{4} \Leftrightarrow$ $\frac{s_l}{l_a+s_l} + \frac{s_l}{l_b+s_l} + \frac{s_l}{l_c+s_l} \ge \frac{9}{4} \Leftrightarrow$ $4*s_l(\frac{1}{l_a+s_l} + \frac{1}{l_b+s_l} + \frac{1}{l_c+s_l}) \ge 9$

Let’s suppose $$l_a \le l_b \le l_c$$, also $$l_a>0$$, then:

$4*s_l(\frac{1}{l_a+s_l} + \frac{1}{l_b+s_l} + \frac{1}{l_c+s_l}) \ge 4*s_l(\frac{1}{l_a+s_l} + \frac{1}{l_a+s_l} + \frac{1}{l_a+s_l}) \ge 9$

So:

$4*s_l(\frac{1}{l_a+s_l}) \ge 3 \Leftrightarrow$ $1-s_l(\frac{1}{l_a+s_l}) \le \frac{1}{4} \Leftrightarrow$ $\frac{l_a}{l_a+l_a+l_b+l_c} \le \frac{l_a}{4*l_a} \le \frac{1}{4}$

$$z=xy^2+x-1$$, and $$z=2xy-1$$.

We can write $$xy^2+x-1=2xy-1$$. Eventually, this relationship becomes equivalent to $$x(y-1)^2$$. But because $$x \neq 0 \Rightarrow y=1$$.

If $$y=1$$, then $$x=\frac{z+y}{2}$$.

We pick a base $$d$$ in the interval of $$a,b,c$$. Basically we as for Problem 18.

After we do the substitution, we need to prove:

$\frac{y+z}{x} + \frac{z+x}{y} + \frac{x+y}{z} \ge \frac{4x}{y+z} + \frac{4y}{z+x} + \frac{4z}{x+y}.$

Then, we need to prove that inequality holds true for:

$\frac{y+z}{x} \ge \frac{4x}{y+z}$

Which is easy to prove using AM-GM inequality.

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