How prove $\sum\frac{1}{2(x+1)^2+1}\ge\frac{1}{3}$ let $x,y,z>0$ and such $xyz=1$ show that
$$\dfrac{1}{2(x+1)^2+1}+\dfrac{1}{2(y+1)^2+1}+\dfrac{1}{2(z+1)^2+1}\ge\dfrac{1}{3}$$
My try: I will find a value of the $k$ such
$$\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{9}+k\ln{x}$$
note $\ln{x}+\ln{y}+\ln{z}=0$,so
$$\sum_{cyc}\dfrac{1}{2(x+1)^2+1}\ge\dfrac{1}{3}+k(\ln{x}+\ln{y}+\ln{z})=\dfrac{1}{3}$$
so let
$$f(x)=\dfrac{1}{2(x+1)^2+1}-k\ln{x}-\dfrac{1}{9}$$
$$\Longrightarrow f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}-\dfrac{k}{x}$$
let $f'(1)=0\Longrightarrow k=-\dfrac{8}{81}$
so
$$f'(x)=\dfrac{-4x-4}{(2x^2+4x+3)^2}+\dfrac{8}{81x}=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}$$
so note when $1>x>\dfrac{1}{2}$ then
$$f'(x)=\dfrac{4(x-1)(8x^3+40x^2+15x-18)}{81x(2x^2+4x+3)^2}<0$$
$x>1,f'(x)>0$
so
$$f(x)\ge f(1)=0$$
so if $x,y,z>\dfrac{1}{2}$ we have prove done.
But for other case,How prove it? Thank you

$z=\dfrac{1}{xy}$, put in LHS and and clean the denominators, we have:
edit:
LHS-RHS=$ 9y^2x^4-8y^3x^3+2y^2x^3+9y^4x^2+2y^3x^2-9y^2x^2-8yx^2-8y^2x+2yx+9 \ge0 \iff $
$4y^2x^4-8yx^2+4\ge 0,\\4y^4x^2-8y^2x+4\ge0,\\5y^2x^4+5y^4x^2\ge 10x^3y^3,\\2y^2x^3+2y^3x^2\ge 4(xy)^{\frac{5}{2}} \iff\\ LHS \ge 2x^3y^3+4(xy)^{\frac{5}{2}}-9x^2y^2+2xy+1=2t^6+4t^5-9t^4+2t^2+1=(t-1)^2(2t^4+8t^3+5t^2+2t+1) \ge0, t=\sqrt{xy}$
the "=" will hold when $xy=1,x=y,y^2x=yx^2=1 \implies x=y=z=1$
