If $a, b, c, d>0$ and $abcd=1$ prove that an inequality holds true If $a, b, c, d>0$ and $abcd=1$ prove that:
$$\frac{a+b+c+d}{4}\ge\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}$$
I attempted to solve it in the following way:
$$\begin{equation}\frac{1}{a^3+b+c+d}+\frac{1}{a+b^3+c+d}+\frac{1}{a+b+c^3+d}+\frac{1}{a+b+c+d^3}\leq\\
\frac{1}{4\sqrt[4]{a^3bcd}}+\frac{1}{4\sqrt[4]{ab^3cd}}+\frac{1}{4\sqrt[4]{abc^3d}}+\frac{1}{4\sqrt[4]{abcd^3}} = \\
\frac{1}{4\sqrt{a}}+\frac{1}{4\sqrt{b}}+\frac{1}{4\sqrt{c}}+\frac{1}{4\sqrt{d}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4\sqrt{abcd}} = \\
\frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4}
\end{equation}$$
This is as far as I got. Could you please help me finish off my thought pattern and finish the question the way I was attempting to solve it?

This is just a thought about the possibility to continue the proof. The inequality:
$$ \frac{a+b+c+d}{4} \ge \frac{\sqrt{bcd}+\sqrt{acd}+\sqrt{abd}+\sqrt{abc}}{4}$$
cannot hold for all $a,b,c,d>0$ with $abcd=1$. In fact, for $N>1$ set $a=b=c=N$, $d=1/N^3$. Then $LHS < N$ and $ RHS > N^{3/2}/4$. Moreover, if $N>16$ then $N^{3/2}/4>N$. Thus, when $N>16$ holds, $LHS<RHS$.
