A Cyclic Inequality in Three Variables XX

By the AM-GM inequality,

$\displaystyle\begin{align}&\sum_{cycl}\sqrt[4]{(a+4b)(2a+3b)(3b+2a)(4a+b)}\\ &\qquad\qquad\le\sum_{cycl}\frac{((a+4b)+(2a+3b)+(3b+2a)+(4a+b)}{4}\\ &\qquad\qquad=5. \end{align}$

Again, by the AM-GM inequality,

$\displaystyle\begin{align}&\sum_{cycl}\sqrt[4]{(a+4b)(2a+3b)(3b+2a)(4a+b)}\\ &\qquad\qquad\ge 5\sum_{cycl}\sqrt[4]{\sqrt[5]{ab^4\cdot a^2b^3\cdot a^3b^2\cdot a^4b}}\\ &\qquad\qquad=5\sum_{cycl}\sqrt{ab}. \end{align}$

This completes the proof.

$\displaystyle \begin{align} &\sqrt[4]{(a+4b)(2a+3b)(3a+2b)(4a+b)}\\ &\qquad\qquad= \sqrt[4]{(4a^2+4b^2+17ab)(6a^2+6b^2+13ab)}\\ &\qquad\qquad\ge \sqrt[4]{(25ab)(25ab)}=5\sqrt{ab}. \end{align}$

Further

$\displaystyle \begin{align} &\sqrt[4]{(a+4b)(2a+3b)(3a+2b)(4a+b)}\\ &\qquad\qquad\le\sum_{cycl}\sqrt[4]{\frac{(a+4b)+(2a+3b)+(3b+2a)+(4a+b)}{4}}\\ &\qquad\qquad=\frac{10}{2}{a+b+c}\\ &\qquad\qquad=5. \end{align}$

This is problem the Romanian Mathematical Magazine, posted by Dan Sitaru at the CutTheKnotMath facebook page. Solution 1 is by Anas Adlany and independently by Diego Alvariz and also by Dang Thanh Tùng; Solution 2 is by Kevin Soto Palacios and independently by Soumava Chakraborty;