Problem. When $a, b, c>0, a, b, c \in \Bbb R, 16(a+b+c)\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, Prove that
$$\sum_{cyc}(\frac{1}{a+b+\sqrt{2a+2c}})^3 \le \frac{8}{9}$$
My approach: If we let $x=a+b, y=b+c, z=c+a$, we can know that $$4(x+y+z) \ge -2\frac{x^2+y^2+z^2-2xy-2yz-2zx}{(-x+y+z)(x-y+z)(x+y-z)}$$
And the inequality that I have to prove will be: $$\sum_{cyc}(\frac{1}{x+\sqrt{2y}})^3 \le \frac{8}{9}$$
But I cannot think further. Can anyone give me a hint?
Let $a=\frac{x}{4},$ $b=\frac{y}{4}$ and $c=\frac{z}{4}.$
Thus, the condition gives $$x+y+z\geq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$ or $$1\leq\frac{xyz(x+y+z)}{xy+xz+yz}$$ and we need to prove that $$\sum_{cyc}\frac{1}{\left(x+y+2\sqrt{2(x+z)}\right)^3}\leq\frac{1}{72}.$$ Now, by AM-GM $$\left(x+y+2\sqrt{2(x+z)}\right)^3\geq\left(3\sqrt[3]{(x+y)\left(\sqrt{2(x+z)}\right)^2}\right)^3=54(x+y)(x+z).$$ Id est, it's enough to prove that $$\sum_{cyc}\frac{1}{(x+z)(y+z)}\leq\frac{3}{4}$$ or $$8(x+y+z)\leq3(x+y)(x+z)(y+z),$$ for which it's enough to prove that $$8(x+y+z)\cdot\frac{xyz(x+y+z)}{xy+xz+yz}\leq3(x+y)(x+z)(y+z),$$ or $$3(x+y)(x+z)(y+z)(xy+xz+yz)\geq8xyz(x+y+z)^2.$$ Now, since $$(x+y)(x+z)(y+z)\geq\frac{8}{9}(x+y+z)(xy+xz+yz)$$ it's $$\sum_{cyc}z(x-y)^2\geq0,$$ it's enough to prove that $$\frac{8}{9}(x+y+z)(xy+xz+yz)\cdot3(xy+xz+yz)\geq8xyz(x+y+z)^2$$ or $$(xy+xz+yz)^2\geq3xyz(x+y+z).$$ Can you end it now?