Unexpectedly real

Algebra Level 3

Let $a,b,c$ be real numbers so that $a+b+c=1$ . Find the maximum value of $\frac { a }{ 1+a^{ 2 } } +\frac { b }{ 1+{ b }^{ 2 } } +\frac { c }{ 1+{ c }^{ 2 } }$ .

If the answer can be written as $\frac{x}{y}$ , where $x,y$ are positive integers so that $\gcd(x,y)=1$ , find $\left| x-y \right|$ .

The answer is 1.

3 solutions

Aaghaz Mahajan
Jun 28, 2018

Relevant wiki: Jensen's Inequality

Consider the function f(x) = $\frac{x}{1+x^2}$

This is concave in (-1,1).......(Can be checked using basic differentiation)
So now, we can use Jensen's inequality to find the maximum which comes out to be $\frac{9}{10}$
Equality holds when a=b=c=1/3......

Well, that's way too short of a solution. Any non-Jensen way?

Steven Jim - 2 years, 11 months ago

@Steven Jim Umm.....well you know as soon as I saw the problem it struck me that Jensen's should be the best way here.......And although it is short, it turns out to be very easy!!! But yeah.....we can try messing around using Titu's lemma I guess.......lemme try it.......

Aaghaz Mahajan - 2 years, 11 months ago

Post a short solution of how you use Jensen anyways :D

Steven Jim - 2 years, 11 months ago

@Steven Jim – Done!!! And yeah we can also use a Non Jensen's approach..... But........I am not really good in creatively using inequalities......what I prefer is Calculus.......(Surprisingly my algebraic method worked in the first try :D !!!)

Aaghaz Mahajan - 2 years, 11 months ago

@Aaghaz Mahajan – Found a way... First, prove that $f\left( a;b;c \right) \le f\left( \frac { a+b }{ 2 } ;\frac { a+b }{ 2 } ;c \right)$ for some configurations of $a$ and $b$ , then substitute $a+b=1-c$ , and boom, one variable. Easy.

Steven Jim - 2 years, 11 months ago

Can we use Jensen's inequality here , after all the concavity changes at root 3 and we have not proved that the maximum is achieved for $a,b,c$ belongs to (-1,1) as you have stated in your solution.

Ankit Kumar Jain - 2 years, 9 months ago

Steven Jim
Jun 28, 2018

Short solution:

First, prove that $f\left( a;b;c \right) \le f\left( \frac { a+b }{ 2 } ;\frac { a+b }{ 2 } ;c \right)$ for some configurations of $a$ and $b$ , then substitute $a+b=1-c$ and the problem is back to one variable.

. .
Feb 19, 2021

Maybe this is the simplest solution.

Since $a + b + c = 1$ , so $| x - y |$ is $\boxed { 1 }$ , too.

What does that mean?

Steven Jim - 3 months, 3 weeks ago

Unexpectedly real

The answer is 1.

3 solutions

0 pending reports