Problem 1612

SOLUTION. WLOG, suppose that $a^{2}+b^{2}+c^{2}=3$. By the weighted Jensen inequality we deduce that
$$\begin{array}{l} \sum_{c y c} \sqrt{\frac{a b}{4 a^{2}+b^{2}+4 c^{2}}}=\sum_{c y c} \frac{a^{2}+4 b^{2}+4 c^{2}}{27} \cdot \sqrt{\frac{27^{2} \cdot a b}{\left(4 a^{2}+b^{2}+4 c^{2}\right)\left(a^{2}+4 b^{2}+4 c^{2}\right)^{2}}} \\ \leq \sqrt{\sum_{c y c} \frac{27 a b}{\left(4 a^{2}+b^{2}+4 c^{2}\right)\left(a^{2}+4 b^{2}+4 c^{2}\right)}}=\sqrt{\sum_{c y c} \frac{3 a b}{\left(4-a^{2}\right)\left(4-b^{2}\right)}} . \end{array}$$

It remains to prove that
$$\begin{array}{c} 3 \sum_{c y c} a b\left(4-c^{2}\right) \leq \prod_{c y c}\left(4-a^{2}\right) \\ \Leftrightarrow 4\left(\sum_{c y c} a b\right)\left(\sum_{c y c} a^{2}\right) \leq \frac{16}{9}\left(\sum_{c y c} a^{2}\right)^{2}+4 \sum_{c y c} a^{2} b^{2}+3 \sum_{c y c} a^{2} b c-a^{2} b^{2} c^{2} \\ \Leftrightarrow 36 \sum_{c y c} a^{3}(b+c)+9 \sum_{c y c} a^{2} b c+9 a^{2} b^{2} c^{2} \leq 16 \sum_{c y c} a^{4}+68 \sum_{c y c} a^{2} b^{2} \end{array}$$

Because $3 a b c \sum_{c y c} a=a b c\left(\sum_{c y c} a\right)\left(\sum_{c y c} a^{2}\right) \geq 9 a^{2} b^{2} c^{2}$, we only need to prove that
$$\begin{array}{l} 9 \sum_{c y c} a^{3}(b+c)+3 \sum_{c y c} a^{2} b c \leq 4 \sum_{c y c} a^{4}+17 \sum_{c y c} a^{2} b^{2} \\ \Leftrightarrow \Leftrightarrow \sum_{c y c}\left(2 a^{2}+2 b^{2}+\frac{3 c^{2}}{2}-5 a b\right)(a-b)^{2} \geq 0 \end{array}$$

Suppose that $a \geq b \geq c$, then we are done by Abel's inequality because $2 b^{2}+2 c^{2}+\frac{3 a^{2}}{2}-5 b c \geq 0$ and
$$\begin{array}{c} \left(2 a^{2}+2 b^{2}+\frac{3 c^{2}}{2}-5 a b\right)+\left(2 a^{2}+2 c^{2}+\frac{3 b^{2}}{2}-5 a c\right)=4 a^{2}+\frac{9}{2}\left(b^{2}+c^{2}\right)-5 a(b+c) \\ \geq 4 a^{2}+9\left(\frac{b+c}{2}\right)^{2}-10 a\left(\frac{b+c}{2}\right) \geq 0 \end{array}$$