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

## Comments

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TopNewestFunctional 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 \( ff(x) = x + 2c \). – Calvin Lin Staff · 2 years, 5 months ago

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– Samuraiwarm Tsunayoshi · 2 years, 5 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.Log in to reply

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. – Calvin Lin Staff · 2 years, 5 months ago

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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. – Akshay Bodhare · 2 years, 5 months ago

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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". – Calvin Lin Staff · 2 years, 5 months ago

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– Samuraiwarm Tsunayoshi · 2 years, 5 months ago

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

The questions are tricky! – Sualeh Asif · 2 years, 5 months ago

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