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

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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,ca,b,c be sides of triangle, s=a+b+c2\displaystyle s = \frac{a+b+c}{2}, prove that for all positive integers nn

anb+c+bnc+a+cna+b(23)n2sn1\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<x1x2x25580 < x_{1} \leq x_{2} \leq \dots \leq x_{2558} such that k=1255811+xk=1\displaystyle \sum\limits_{k=1}^{2558} \frac{1}{1+x_{k}} = 1, prove that

k=12558xk2557(k=125581xk)\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:RRf: \mathbb{R} \rightarrow \mathbb{R} that satisfy for all x,y,zRx,y,z \in \mathbb{R}.

f(xy)2+f(xz)2f(x)f(yz)14\displaystyle \frac{f(xy)}{2}+\frac{f(xz)}{2}-f(x)f(yz) \geq \frac{1}{4}

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

f(x)+f1(x)=2xf(x)+f^{-1}(x) = 2x

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

Note by Samuraiwarm Tsunayoshi
4 years, 6 months ago

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Functional Equations 1) Hints

Step 1: Find f(0) f(0) .
Step 2: Show that for f(yz)f(y) 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 f(x) = x+c , then ff(x)=x+2c ff(x) = x + 2c .

Calvin Lin Staff - 4 years, 6 months ago

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

Samuraiwarm Tsunayoshi - 4 years, 6 months ago

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

Let x=0,y=0,z=0 x = 0, y = 0, z = 0 , we obtain f(0)(1f(0)14 f(0) ( 1 - f(0 ) \geq \frac{1}{4} , hence we can conclude that f(0)=12 f(0) = \frac{1}{2} .

Let x=x,0y=y,z=0 x = x, \neq 0 y = y, z = 0 , we obtain f(xy)2+14f(x)×1214 \frac{ f(xy) } { 2} + \frac{1}{4} - f(x) \times \frac{1}{2} \geq \frac{1}{4} , or that f(xy)f(x) f(xy) \geq f(x) .
But since this is true for all y y , with x=x,y=yx x = x, y = \frac{y}{x} , we obtain f(y)f(x) f( y) \geq f(x) for arbitrary x,yx, y .. Hence, the only possible function is a constant function, except possibly for x=0 x = 0 .

Now, for any x0 x \neq 0 , let x=y=z x = y = z , and set f(x)=a f(x) = a . We get aa214 a - a^2 \geq \frac {1}{4} , from which we can conclude that a=12 a = \frac{1}{2} .

Thus, the only solution is f(x)=12x f(x) = \frac{1}{2} \forall x .


Note: Yes, it is easy to find f(1)f(1) by setting x=y=z=1 x = y = z = 1 to obtain f(1)=12 f(1) = \frac{1}{2} . There are likely alternative approaches which uses this fact.

Calvin Lin Staff - 4 years, 6 months ago

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This isn't a full solution.

For first inequality,

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

Step 2) The denominator will be in terms of 1an+1bn+1cn\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}.Minimize this using Hölder's inequality to the form of 1a+1b+1c\frac{1}{a}+\frac{1}{b}+\frac{1}{c},then again using AMHMAM\geq HM write it in terms of ss 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)(a=b=c). Finally,we arrive at the desired answer.

Akshay Bodhare - 4 years, 6 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 - 4 years, 6 months ago

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

Samuraiwarm Tsunayoshi - 4 years, 6 months ago

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Can you give solutions especially the functional equations part,

The questions are tricky!

Sualeh Asif - 4 years, 6 months ago

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