An Inequality with Integrals

Problem

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.

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Leo Giugiuc's Cyclic Inequality with a Constraint $\left(\displaystyle a^3+b^3+c^3+\frac{16}{(a+b)(b+c)(c+a)}\ge 5\right)$

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