How prove this inequality $\frac{1}{(a+1)^2+\sqrt{2(b^4+1)}}+\frac{1}{(b+1)^2+\sqrt{2(c^4+1)}}+\frac{1}{(c+1)^2+\sqrt{2(a^4+1)}}\le\frac{1}{2}$

Question

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

Answer 1

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