Which one should I apply?

Algebra Level 3

$\large \dfrac{1}{c^2+a+b}+\dfrac{1}{b^2+a+c}+\dfrac{1}{a^2+b+c} \; .$

If $a,b$ and $c$ are positive numbers satisfying $a+b+c=3$ , find maximum value of the expression above.

The answer is 1.

2 solutions

Bolin Chen
Mar 6, 2016

Use Cauchy Inequality, $(c^2+a+b)(1+a+b)$ ≥ $(c+a+b)^2$ =9

So, $\frac{1}{c^2+a+b}$ + $\frac{1}{b^2+a+c}$ + $\frac{1}{a^2+b+c}$ ≤ $\frac{1+a+b}{9}$ + $\frac{1+b+c}{9}$ + $\frac{1+a+c}{9}$ = $\frac{3+2(a+b+c)r}{9}$ =1

Q.E.D.

You can use AM-HM inequality.

Venkata Karthik Bandaru - 5 years, 3 months ago

In fact,there is another solution useing Chebyshev's Inequality:)

Bolin Chen - 5 years, 3 months ago

Indraneel Mukhopadhyaya
Mar 6, 2016

Consider the function $f(x)=\dfrac{1}{(x^2-x+3)}$ for $x>0$ .We have to find the maximum value of $f(a)+f(b)+f(c)$ .Note that $f''$ is negative for $x>0$ .Hence using Jensen's inequality the maximum value of the given expression is $3\times f\left(\dfrac{a+b+c}{3}\right)=\boxed{1}$

Nice solution~

Bolin Chen - 5 years, 3 months ago

Which one should I apply?

The answer is 1.

2 solutions

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