TITLE: prove $\frac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geqslant \frac{3}{\sqrt[4]{2}}$
QUESTION [11 upvotes]: $a,b,c >0$ and $abc=1$, prove
$$\frac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geqslant \frac{3}{\sqrt[4]{2}}$$
1. I tried rearrangement and AM-GM but fail.
2. I think the power of $\frac14$ is tough. I can prove the easier inequality 
$$\frac{1}{a^3(a+b^2)}+\frac{1}{b^3(b+c^2)}+\frac{1}{c^3(c+a^2)} \geqslant \frac{3}{2} $$

REPLY [2 votes]: From AM–GM inequality

$\dfrac{\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\dfrac{1}{\sqrt[4]{b^3(b+c^2)}}+\dfrac{1}{\sqrt[4]{c^3(c+a^2)}}}{3} \geq \sqrt[3]{\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}\times\frac{1}{\sqrt[4]{c^3(c+a^2)}}}$
$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geq 3\sqrt[3]{\dfrac{1}{\sqrt[4]{a^3(a+b^2)b^3(b+c^2)c^3(c+a^2)}}}$
$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}}
 \geq 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(abc)^3(a+b^2)(b+c^2)(c+a^2)}}}$
$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\frac{1}{\sqrt[4]{b^3(b+c^2)}}+\frac{1}{\sqrt[4]{c^3(c+a^2)}} \geq 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}} \cdots(1)$

From AM–GM inequality

$\dfrac{(a+b^2)+(b+c^2)+(c+a^2)}{3}\ge \sqrt[3]{(a+b^2)(b+c^2)(c+a^2)}$
$\dfrac{(a+b+c)+(a^2+b^2+c^2)}{3}\ge \sqrt[3]{(a+b^2)(b+c^2)(c+a^2)} 
> \cdots(2)$

From AM–GM inequality

$\dfrac{a+b+c}{3}\ge \sqrt[3]{abc}$
$\dfrac{a+b+c}{3}\ge \sqrt[3]{1}$
$a+b+c \ge 3 \cdots(3)$

From AM–GM inequality

$\dfrac{a^2+b^2+c^2}{3}\ge \sqrt[3]{a^2b^2c^2}$
$\dfrac{a^2+b^2+c^2}{3}\ge \sqrt[3]{(abc)^2}$
$\dfrac{a^2+b^2+c^2}{3}\ge 1$
$a^2+b^2+c^2 \ge 3 \cdots(4)$

Consider the equations $(3)$ and $(4)$. From these, we have

$(a+b+c)+(a^2+b^2+c^2) \ge 3+3$
$(a+b+c)+(a^2+b^2+c^2) \ge 6$
$\dfrac{(a+b+c)+(a^2+b^2+c^2)}{3} \ge 2 \cdots(5)$

Consider the equations $(2)$ and $(5)$. From these, we have

$2 \ge \sqrt[3]{(a+b^2)(b+c^2)(c+a^2)}$
$8 \ge (a+b^2)(b+c^2)(c+a^2)$
$\sqrt[4]{8} \ge \sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}$
$\dfrac{1}{\sqrt[4]{8}} \le
 \dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}$
$\left(\dfrac{1}{\sqrt[4]{8}}\right)^{1/3} \le
 \sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}}$
$3\left(\dfrac{1}{\sqrt[4]{8}}\right)^{1/3} \le
 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}}$
$\dfrac{3}{\left(\sqrt[4]{8}\right)^{1/3}} \le
 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}}$
$\dfrac{3}{\sqrt[4]{2}} \le 
 3\sqrt[3]{\dfrac{1}{\sqrt[4]{(a+b^2)(b+c^2)(c+a^2)}}} ~~\cdots(6)$

Consider the equations $(1)$ and $(6)$. From these, we have

$\dfrac{1}{\sqrt[4]{a^3(a+b^2)}}+\dfrac{1}{\sqrt[4]{b^3(b+c^2)}}+\dfrac{1}{\sqrt[4]{c^3(c+a^2)}}
 \ge \dfrac{3}{\sqrt[4]{2}}$