# An Inequality with Five Variables, Only Three Cyclic

### Solution 1

With the power-mean inequality, $\displaystyle \frac{1}{n}\sum_{k=1}^nx_k^s\ge\left(\frac{1}{n}\sum_{k=1}^nx_k\right)^s,$

$\displaystyle \left(a+\frac{b}{c}\right)^4+ \left(a+\frac{b}{d}\right)^4+ \left(a+\frac{b}{e}\right)^4\ge\frac{1}{27}\left(3a+\frac{b}{c}+\frac{b}{d}+\frac{b}{e}\right)^4.$

Thus, suffice it to prove that

$\displaystyle \frac{1}{27}\left(3a+\frac{b}{c}+\frac{b}{d}+\frac{b}{e}\right)^4\ge 3(a+3b)^4.$

This is equivalent to

$\displaystyle 3a+\frac{b}{c}+\frac{b}{d}+\frac{b}{e}\ge 3(a+3b),$

or,

$\displaystyle \frac{1}{c}+\frac{1}{d}+\frac{1}{e}\ge 9.$

On the other hand, by Hölder's inequality,

$\displaystyle (c+d+e)\left(\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right)\ge (1+1+1)^2=9.$

Now, since $c+d+e=1,$ $\displaystyle \frac{1}{c}+\frac{1}{d}+\frac{1}{e}\ge 9.$

### Solution 2

In so far as $a,b\gt 0,$ function $\displaystyle f(x)=\left(a+\frac{b}{x}\right)^4$ is convex for $x\gt 0,$, so that, by Jensen's inequality,

$\displaystyle \frac{1}{3}\left(f(c)+f(d)+f(e)\right)\ge f\left(\frac{c+d+e}{3}\right)=f\left(\frac{1}{3}\right).$

In other words,

$\displaystyle \frac{1}{3}\left[\left(a+\frac{b}{c}\right)^4+ \left(a+\frac{b}{d}\right)^4+ \left(a+\frac{b}{e}\right)^4\right]\ge \left(a+\frac{b}{1/3}\right)^4=(a+3b)^4.$

### Solution 3

\displaystyle\begin{align} \left(a+\frac{b}{c}\right)^4+\left(a+\frac{b}{d}\right)^4+\left(a+\frac{b}{e}\right)^4 &\geq 3\left[a+\frac{b}{3}\left(\frac{1}{c}+\frac{1}{d}+\frac{1}{e}\right)\right]^4~\text{(Jensen's)}\\ &\geq 3\left[a+\frac{b}{3}\left(\frac{9}{c+d+e}\right)\right]^4~\text{(AM-HM)} \\ &=3(a+3b)^4. \end{align}

### Acknowledgment

Dan Sitaru has kindly posted the problem at the CutTheKnotMath facebook page. Additional solutions can be found at the link.

Solution 2 is by Long Huynh Huu; Solution 3 is by Amit Itagi.