<s>[INST] 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
 [/INST] Hint:
Use Tringnometric transformation such that 
x=sinu, y=secu and z = cotu
For any value of u you can find x,y,z >0 such that xyz = 1 by definition.
For a value of u = pi/6, x=.5, y = 1.154 and z = 1.732
Further, I computed LHS, for u = pi/4 and its value is .3365
Just to double check, I computed LHS for u = pi/3 and its value is .34552
In both sample cases, LHS is >= (1/3).
Thanks
Satish
</s>