# Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots II

### Solution 1

By the AM-GM inequality,

\displaystyle \begin{align} 1&=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{1}{3}\sum_{cycl}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\\ &\ge \frac{1}{3}\cdot 3\cdot\sum_{cycl}\frac{1}{\sqrt[3]{abc}}=\sum_{cycl}\frac{1}{\sqrt[3]{abc}}. \end{align}

Thus,$\displaystyle \sum_{cycl}\frac{1}{\sqrt[3]{abc}}\le 1.$ Multiplying this by $\sqrt[3]{abcd}$ yields the required inequality.

Equality is attained for $a=b=c=d=4.$

### Solution 2

Due to the constraint,

$\displaystyle 1=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)^3\ge 4^2\left(\frac{1}{abc}+\frac{1}{bcd}+\frac{1}{cda}+\frac{1}{dab}\right),$

implying

$\displaystyle \frac{1}{abc}+\frac{1}{bcd}+\frac{1}{cda}+\frac{1}{dab}\le\frac{1}{4^2}.$

Hence, $\displaystyle 4^2(a+b+c+d)\le abcd,\,$ so that $\displaystyle \sqrt[3]{4^2(a+b+c+d)}\le \sqrt[3]{abcd},$ but (due to concavity of $y=\sqrt[3]{x},$ for $x\ge 0,$ and Jensen's inequality) $\displaystyle \sum_{cycl}\sqrt[3]{a}\le\sqrt[3]{4^2\sum_{cycl}a},$ such that, finally, $\displaystyle \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+\sqrt[3]{d}\le\sqrt[3]{abcd}.$

### Solution 3

Let $\displaystyle \frac{1}{a}=p^3,~\frac{1}{b}=q^3,~\frac{1}{c}=r^3,~\frac{1}{d}=s^3.$ From Muirhead (or four AM-GM's taking three cubes at a time),

$\displaystyle pqr+qrs+rsp+spq\leq p^3+q^3+r^3+s^3=1$.

Dividing both sides by $pqrs$, we get

$\displaystyle \frac{1}{p}+\frac{1}{q}+\frac{1}{r}+\frac{1}{s}\leq \frac{1}{pqrs},$

which in terms of $a-d$ is the inequality to be proved.

### Solution 4

We have the Harmonic Mean $\displaystyle \frac{\displaystyle 4}{\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}=4$, so by the AM-HM inequality, $a+b+c+d \geq 16.$ Also, by the AM-GM inequality,

$\displaystyle 1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq 4 \sqrt[4]{\frac{1}{a b c d}}.$

Therefore, $\displaystyle \sqrt[3]{a b c d}\geq 4\times 2^{2/3}.$ So,we have (by the Power-Mean inequality)

$\displaystyle \frac{1}{64} \left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+\sqrt[3]{d}\right)^3\leq \frac{1}{4} (a+b+c+d)\leq 4.$

Finally

$\displaystyle \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+\sqrt[3]{d}\leq 4\times 2^{2/3}\le\sqrt[3]{a b c d}.$

### Acknowledgment

Dan Sitaru has kindly posted the problem at the CutTheKnotMath facebook page, with solutions by Yadamsuren Myagmarsuren (Solution 1) and Sanong Hauerai (Solution 2). The problem was originally publshed at the Romanian Mathematical Magazine. Solution 3 is by Amit Itagi; Solution 4 is by N. N. Taleb.