# Inequalities & Functional Equations (Thailand Math POSN 3rd round)

Write a full solution.

Inequalities

Theorems allowed to use: basic inequalities, AM-GM-HM, Cauchy Schwarz, Holder, Weighted AM-GM, Jensen, Power Mean, Chebyshev, Rearrangement, Weirstrass, Bernoulli

Comments by professor: These 2 problems can be solved using AM-GM and Cauchy Schwarz only. Maybe you can try it!

1.) Let $$a,b,c$$ be sides of triangle, $$\displaystyle s = \frac{a+b+c}{2}$$, prove that for all positive integers $$n$$

$\displaystyle \frac{a^{n}}{b+c}+\frac{b^{n}}{c+a}+\frac{c^{n}}{a+b} \geq \left(\frac{2}{3}\right)^{n-2}s^{n-1}$

2.) Let $$0 < x_{1} \leq x_{2} \leq \dots \leq x_{2558}$$ such that $$\displaystyle \sum\limits_{k=1}^{2558} \frac{1}{1+x_{k}} = 1$$, prove that

$\displaystyle \sum\limits_{k=1}^{2558}\sqrt{x_{k}} \geq 2557\left(\sum\limits_{k=1}^{2558}\frac{1}{\sqrt{x_{k}}}\right)$

Functional Equations

1.) Find all functions $$f: \mathbb{R} \rightarrow \mathbb{R}$$ that satisfy for all $$x,y,z \in \mathbb{R}$$.

$\displaystyle \frac{f(xy)}{2}+\frac{f(xz)}{2}-f(x)f(yz) \geq \frac{1}{4}$

2.) Find all strictly monotonic functions $$f: \mathbb{R} \rightarrow \mathbb{R}$$ that satisfy for all $$x \in \mathbb{R}$$.

$f(x)+f^{-1}(x) = 2x$

This note is part of Thailand Math POSN 3rd round 2015

Note by Samuraiwarm Tsunayoshi
3 years, 7 months ago

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Functional Equations 1) Hints

Step 1: Find $$f(0)$$.
Step 2: Show that for $$f(yz) \geq f(y)$$.

Functional Equations 2) Hints
Step 1: Guess what all the possible functions are.
Step 2: Show that if $$f(x) = x+c$$, then $$ff(x) = x + 2c$$.

Staff - 3 years, 7 months ago

For 1) you can find that $$f(x)$$ is bounded above by some value and bounded below by some value. You need to find $$f(1)$$ though.

- 3 years, 7 months ago

There are likely multiple ways of approaching these functional equations. Here's what i did for 1, without having to find $$f(1)$$ explicitly.

Let $$x = 0, y = 0, z = 0$$, we obtain $$f(0) ( 1 - f(0 ) \geq \frac{1}{4}$$, hence we can conclude that $$f(0) = \frac{1}{2}$$.

Let $$x = x, \neq 0 y = y, z = 0$$, we obtain $$\frac{ f(xy) } { 2} + \frac{1}{4} - f(x) \times \frac{1}{2} \geq \frac{1}{4}$$, or that $$f(xy) \geq f(x)$$.
But since this is true for all $$y$$, with $$x = x, y = \frac{y}{x}$$, we obtain $$f( y) \geq f(x)$$ for arbitrary $$x, y$$.. Hence, the only possible function is a constant function, except possibly for $$x = 0$$.

Now, for any $$x \neq 0$$, let $$x = y = z$$, and set $$f(x) = a$$. We get $$a - a^2 \geq \frac {1}{4}$$, from which we can conclude that $$a = \frac{1}{2}$$.

Thus, the only solution is $$f(x) = \frac{1}{2} \forall x$$.

Note: Yes, it is easy to find $$f(1)$$ by setting $$x = y = z = 1$$ to obtain $$f(1) = \frac{1}{2}$$. There are likely alternative approaches which uses this fact.

Staff - 3 years, 7 months ago

This isn't a full solution.

For first inequality,

Step 1)Use $$AM\geq HM$$ after writing $$b+c=2s-a$$

Step 2) The denominator will be in terms of $$\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}$$.Minimize this using Hölder's inequality to the form of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$,then again using $$AM\geq HM$$ write it in terms of $$s$$ and substitute this minimized value in original denominator,but we should careful while substituting such inequalities.Thus,we need to check equality condition for each and every step.However,it is same for each step$$(a=b=c)$$. Finally,we arrive at the desired answer.

- 3 years, 7 months ago

Nice set of problems.

For question 2, the initial assumption of "ordered chain" isn't necessary, since the LHS and RHS are both independent of the order of the sequence. I wonder why that put that condition in, rather than simply saying "all positive".

Staff - 3 years, 7 months ago

I think that the condition gives a hint for Chebyshev's Inequality. Not really sure though.

- 3 years, 7 months ago

Can you give solutions especially the functional equations part,

The questions are tricky!

- 3 years, 7 months ago