An Inequality with Integrals

Problem

An Inequality with Integrals

Solution 1

$\displaystyle \begin{align} \Omega(u)&=\int_0^u\frac{x^4+7x^3-25x^2+37x+4}{x^4-3x^3+5x^2-3x+4}dx\\ &=\int_0^u\frac{(x^4-3x^3+5x^2-3x+4)+(10x^3-30x^2+40x)}{x^4-3x^3+5x^2-3x+4}dx\\ &=\int_0^u\left(1+\frac{10x(x^2-3x+4)}{(x^2+1)(x^2-3x+4)}\right)dx\\ &=\int_0^u\left(1+\frac{10x}{x^2+1}\right)dx\\ &=u+5\ln(u^2+1). \end{align}$

It follows that, by the AM-GM inequality, that

$\displaystyle \begin{align} \Omega(a)+\Omega(b)+\Omega(C)&\ge (a+b+c)+5\sum_{cycl}\ln(a^2+1)\\ \Omega(a)+\Omega(b)+\Omega(C)&\ge 3\sqrt[3]{abc}+5\ln\prod_{cycl}(a^2+1)\\ &=3+5\ln\prod_{cycl}(a^2+1). \end{align}$

Solution 2

$\displaystyle \Omega (u):=\int_0^u f(x)\, dx$ where

$\displaystyle f(x)=\frac{x^4+7 x^3-25 x^2+37 x+4}{x^4-3 x^3+5 x^2-3 x+4}.$

Simplifying we get $\displaystyle f(x)=\frac{10 x}{x^2+1}+1.$

We have the antiderivative as

$\displaystyle \int f(x)\, dx= 5 \log \left(x^2+1\right)+x.$

$\displaystyle\sum_{cycl}(\Omega)=\sum_{cycl}a+ 5\log \left(x^2+1\right).$ Since $a+b+c \geq 3 \sqrt[3]{a b c}=3$, by AM-GM, we get the required proof.

Acknowledgment

The problem was kindly posted at the CutTheKnotMath facebook page by Dan Sitaru. The problem is by Dan and was previously published in the Romanian Mathematical Magazine. Solution 1 is by Seyran Ibrahimov; Solution 2 is by N. N. Taleb.

 

Inequalities with the Product of Variables as a Constraint

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