<s>[INST] Prove that : $\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a} \le 3\sqrt[4]{2}.$ For $a,b,c\in\mathbb{R}^+$ and $a+b+c=3$ . 
Prove that : $\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a} \le 3\sqrt[4]{2}.$
 [/INST] Oops, give a wrong answer in previous edit. Here is the corrected version.
Notice both $\sqrt[4]{x}$ and $\sqrt[4]{x^2+1}$ are strictly increasing function in $x$.
The list of numbers $(\sqrt[4]{a},\sqrt[4]{b},\sqrt[4]{c})$ are in same sorted order as $(\sqrt[4]{a^2+1},\sqrt[4]{b^2+1},\sqrt[4]{c^2+1})$.
By Rearrangement inequality, we have:
$$\begin{align}&\sqrt[4]{(a^2+1)b}+\sqrt[4]{(b^2+1)c}+ \sqrt[4]{(c^2+1)a}\\
\le & \sqrt[4]{(a^2+1)a}+\sqrt[4]{(b^2+1)b}+ \sqrt[4]{(c^2+1)c}\tag{*1}
\end{align}$$
Notice
$$\frac{d^2}{dx^2} \sqrt[4]{(x^2+1)x} = 
-\frac{3\,{\left( x^2 -1\right) }^{2}}{16\,{x}^{\frac{7}{4}}\,{\left( {x}^{2}+1\right) }^{\frac{7}{4}}} \le 0$$
$\sqrt[4]{(x^2+1)x}$ is a concave function in $x$ and by Jensen's inequality, we have:
$$
\text{R.H.S of } (*1)
\le 3 \left[ \left( \left(\frac{a+b+c}{3}\right)^2 + 1\right) \left(\frac{a+b+c}{3}\right)\right]^{\frac14}
= 3\sqrt[4]{2}
$$
</s>