# A Cyclic Inequality in Three Variables X

### Solution 1

First note that from AM-QM inequality, $\displaystyle\left(\frac{a+b+c}{3}\right)^2\le\frac{a^2+b^2+c^2}{3}=1\,$ so that $a+b+c\le 3.$

Now, both functions, $\displaystyle y=\frac{1}{(x+1)^3}\,$ and $\displaystyle y=\frac{1}{(x+1)^4}\,$ are convex on $(0,\infty),\,$ so that, by Jensen's inequality

$\displaystyle\sum_{cycl}\frac{1}{(a+1)^3}\ge\frac{3}{\left(\displaystyle\frac{a+b+c}{3}+1\right)^3}\ge\frac{3}{8}\,$ and
$\displaystyle\sum_{cycl}\frac{4}{(a+1)^4}\ge\frac{3\cdot 4}{\left(\displaystyle\frac{a+b+c}{3}+1\right)^4}\ge\frac{3}{4}.\,$

### Solution 2

We obtain the same result by using Radon's inequality:

$\displaystyle\sum_{cycl}\frac{1}{(a+1)^3}\ge\frac{(1+1+1)^4}{(a+b+c+3)^3}\ge\frac{3}{8}\,$ and
$\displaystyle\sum_{cycl}\frac{4}{(a+1)^4}\ge\frac{4(1+1+1)^5}{(a+b+c+3)^4}\ge\frac{3}{4}.\,$

### Solution 3

Define $\displaystyle f(x)=\frac{1}{(x+1)^3}+\frac{4}{(x+1)^4}.\,$ We have

$\displaystyle f(x)-\left(\frac{17}{16}-\frac{11x}{16}\right)=\frac{(x-1)^2(11x^3+49x^2+85x+63)}{16(x+1)^4}\ge 0,$

implying

$\displaystyle\sum_{cycl}f(a)\ge\sum_{cycl}\left(\frac{17}{16}-\frac{11a}{16}\right)=3\cdot\frac{17}{16}-(a+b+c)\cdot\frac{11}{16}.$

But from $\displaystyle\sum_{cycl}a^2=3\,$ it follows that $a+b+c\le 3.\,$ Thus,

$\displaystyle\sum_{cycl}f(a)\ge\frac{51}{16}-\frac{33}{16}=\frac{9}{8}.$

### Solution 4

Let define $x=a^2,\,y=b^2,\,z=c^2.\,$ The problem becomes

Let $x,y,z\gt 0\,$ satisfy $x+y+z=3.\,$ Prove that

$\displaystyle\sum_{cycl}\frac{\sqrt{x}+5}{(\sqrt{x}+1)^4}\ge\frac{9}{8}.$

Note that the function $\displaystyle f(u)=\frac{\sqrt{u}+5}{(\sqrt{u}+1)^4}\,$ is convex on $(0,\infty)\,$ and thus the problem lends itself to Jensen's inequality:

$\displaystyle\sum_{cycl}\frac{\sqrt{x}+5}{(\sqrt{x}+1)^4}\ge 3\frac{\displaystyle\sqrt{\frac{3}{3}}+5}{\displaystyle\left(\sqrt{\frac{3}{3}}+1\right)^4}=\frac{9}{8}.$

### Acknowledgment

Dan Sitaru has kindly posted the above problem (from his book "Math Accent") at the CutTheKnotMath facebook page to which Leo Giugiuc responded with Solution 1. Solution 3 is by Imad Zak; Solution 4 is by Amit Itagi.