# An Inequality with Arbitrary Roots

### Solution 1

By the AM-GM inequality,

\displaystyle \begin{align} \sqrt[n]{a+\sqrt[n]{a}} &= \sqrt[n]{a^{\frac{n-1}{n}}\left(a^{\frac{1}{n}}+a^{-\frac{n-2}{n}}\right)}\\ &\le\frac{\displaystyle (n-1)a^{\frac{1}{n}}+a^{\frac{1}{n}}+a^{-\frac{n-2}{n}}}{n}\\ &=a^{\frac{1}{n}}+\frac{1}{n}a^{-\frac{n-2}{n}}. \end{align}

Similarly,

$\displaystyle \sqrt[n]{a-\sqrt[n]{a}}\le a^{\frac{1}{n}}-\frac{1}{n}a^{-\frac{n-2}{n}}$

so that

(1)

$\displaystyle \sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\le 2\sqrt[n]{a}.$

On the other hand, again by the AM-GM inequality, $\displaystyle \sqrt[n]{a\left(3^n\right)^{n-1}}\le \frac{a+(n-1)3^n}{n},\,$ implying

(2)

$\displaystyle \sqrt[n]{a} \le \frac{a+(n-1)3^n}{n\sqrt[n]{\left(3^n\right)^{n-1}}}=\frac{a+(n-1)3^n}{n3^{n-1}}.$

From (1) and (2),

$\displaystyle \sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\le \frac{2a+(n-1)3^n}{n3^{n-1}}.$

Thus

\displaystyle\begin{align} \sum_{cycl}\left(\sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\right)&\le \sum_{cycl}\frac{\displaystyle 2a+(n-1)3^{n}}{\displaystyle n3^{n-1}}\\ &=\frac{\displaystyle 2\sum_{cycl}a+(n-1)3^{n+1}}{\displaystyle n3^{n-1}}\\ &=\frac{\displaystyle 2\cdot 3^{n+1}+(n-1)3^{n+1}}{\displaystyle n3^{n-1}}\\ &=\frac{(n+1)3^{n+1}}{n3^{n-1}}\\ &=\frac{9(n+1)}{n}\lt 18. \end{align}

### Solution 2

By the $m$th power theorem, with $\displaystyle m=\frac{1}{n},\,$ we get

\displaystyle\begin{align} \frac{\sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}}{2}&\le \left(\frac{a+\sqrt[n]{a}+a-\sqrt[n]{a}}{2}\right)^{\frac{1}{n}}\\ &=\sqrt[n]{a}, \end{align}

so that

$\displaystyle \sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\le 2\sqrt[n]{a}.$

Summing up,

\displaystyle\begin{align} \sum_{cycl}\left(\sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\right)&\le 2\sum_{cycl}\sqrt[n]{a}\\ &\le 2\cdot 3\left(\frac{\displaystyle \sum_{cycl}a}{3}\right)^{\frac{1}{n}}\\ &=6\left(\frac{3^{n+1}}{3}\right)^{\frac{1}{n}}=6\cdot 3=18. \end{align}

### Solution 3

From power-mean inequality,

$2\sqrt[n]{a} > \left(\sqrt[n]{a+\sqrt[n]{a}}+\sqrt[n]{a-\sqrt[n]{a}}\right), ~\text{(and its cyclic variants)}$

and

$\displaystyle 6\sqrt[n]{\frac{a+b+c}{3}} = 18 \geq 2\left(\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}\right).$

Putting the two together completes the proof. Note that the first inequality has $\gt\,$ and not $\geq\,$ because the two terms in the sum cannot be equal.

### Solution 4

By concavity of $(.)^{\frac{1}{n}}$ with $n\gt 2$, by Jensen's inequality,

$\displaystyle \frac{1}{2}\left(a-a^{1/n}\right)^{1/n}+\frac{1}{2}\left(a+a^{1/n}\right)^{1/n}\leq a^{1/n}$

Since $\displaystyle \left(\frac{1}{3} \left(a^{1/n}+b^{1/n}+c^{1/n}\right)\right)^n\leq \frac{1}{3} (a+b+c),\,$ we have

$\displaystyle \text{lhs}\leq 2 \left(a^{1/n}+b^{1/n}+c^{1/n}\right)\leq 2 \left(3^{n-1} 3^{n+1}\right)^{1/n}=18$

### Acknowledgment

Dan Sitaru has kindly posted the problem of his from the Romanian Mathematical Magazine at the CutTheKnotMath facebook page. Solution 1 is by Khanh Hung Vu; Solution 2 is by Abdur Rahman; Solution 3 is by Amit Itagi; Solution 4 is by N. N. Taleb.