How prove this inequality $\frac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\frac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\frac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\frac{1}{2}$ let  $a,b,c>0$,and such $abc=1$, show that
$$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\dfrac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\dfrac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\dfrac{1}{2}$$
My idea : Use Cauchy-Schwarz inequality,we have
$$2(b^4+1)=(1+1)(b^4+1)\ge (b^2+1)^2$$
so
$$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}\le\dfrac{1}{(a+1)^2+(b^2+1)}\le\dfrac{1}{(a+1)^2+\dfrac{(b+1)^2}{2}}$$
then we only prove
$$\Longleftrightarrow\sum_{cyc}\dfrac{2}{2(a+1)^2+(b+1)^2}\le\dfrac{1}{2}$$
$$\Longleftrightarrow\sum_{cyc}\dfrac{1}{2(a+1)^2+(b+1)^2}\le\dfrac{1}{4}$$
Then I can't.Thank you

since Use Cauchy-Schwarz inequality and AM-GM inequality,we have
$$\dfrac{1}{(a+1)^2+\sqrt{2(b^4+1)}}\le\dfrac{1}{(a+1)^2+b^2+1}\le\dfrac{1}{2ab+2a+2}=\dfrac{1}{2}\cdot\dfrac{1}{ab+a+1}$$
Use  follow well know reslut,if $abc=1$,then we have
$$\sum_{cyc}\dfrac{1}{ab+a+1}=1$$
poof: since
\begin{align*}\dfrac{1}{ab+a+1}+\dfrac{1}{bc+b+1}+\dfrac{1}{ac+c+1}&=\dfrac{1}{ab+a+1}+\dfrac{a}{abc+ab+a}+\dfrac{ab}{abc\cdot a+abc+1}\\
&=\dfrac{ab+a+1}{ab+a+1}\\
&=1
\end{align*}
so the inequality we have By Done
