A Little More of Algebra for an Inequality
A Little Less of Calculus for a Generalization

Source

Dorin Marghidanu's Inequality with Powers and Reciprocals, source

Problem

Let $a,b,c$ be positive real numbers. Prove that

$\displaystyle \sum_{cycl}\frac{a}{a^2bc+b^4+c^4}\le\frac{1}{abc}.$

Solution 1

For real $b,c,$ $(b-c)(b^3-c^3)\ge 0,$ implying $b^4+c^4\ge bc^3+b^3c=bc(b^+c^2),$ with equality only if $b=c.$ Derive similarly two additional inequalities. It follows that

$\displaystyle\begin{align} \sum_{cycl}\frac{a}{a^2bc+b^4+c^4}&\le\sum_{cycl}\frac{a}{bc(a^2+b^2+c^2)}\\ &=\sum_{cycl}\frac{a^2}{abc(a^2+b^2+c^2)}\\ &=\frac{1}{abc}\cdot\frac{a^2+b^2+c^2}{a^2+b^2+c^2}\\ &=\frac{1}{abc}, \end{align}$

with equality only if $a=b=c.$

Solution 2

Note that $bc(b^2+c^2)\le b^4+c^4,$ due to Muirhead's inequality. Then we follow Solution 1.

Solution 3

Let $abc=1.$ We shall prove that

$\displaystyle \sum_{cycl}\frac{a}{a+b^4+c^4}\le 1,$

which is equivalent to

$\displaystyle \sum_{cycl}\frac{b^4+c^4}{a+b^4+c^4}\ge 2,$

With the Cauchy-Schwarz inequality, we have

$\displaystyle \begin{align} &\sum_{cycl}\frac{b^4+c^4}{a+b^4+c^4}\ge\frac{\left(\sqrt{a^4+b^4}+\sqrt{b^4+c^4}+\sqrt{c^4+a^4}\right)^2}{2(a^4+b+4+c^4)+a+b+c}\\ &\qquad=\frac{2(a^4+b^4+c^4)+2\sum_{cycl}\sqrt{(a^4+b^4)(b^4+c^4)}}{2(a^4+b+4+c^4)+a+b+c}\\ &\qquad\ge\frac{2(a^4+b^4+c^4)+2(a^4+b^4+c^4+a^2b^2+b^2c^2+c^2a^2)}{2(a^4+b+4+c^4)+a+b+c}\\ &\qquad\ge\frac{{4(a^4+b^4+c^4)+2(a+b+c)}}{2(a^4+b+4+c^4)+a+b+c}=2. \end{align}$

Taking into account $abc=1,$ the last inequality follows by rearrangement from

$\displaystyle \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c.$

Acknowledgment

Dorin Marghidanu has kindly posted the problem at the CutTheKnotMath facebook page, along with his solution (Solution 1) and later commented with a link to additional solutions. Solution 2 is by Leo Giugiuc; Solution 3 is by Nguyễn Ngọc Tú

 

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