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?
 A: We can not finish it because the inequality, which you'll get after your step is wrong.
Indeed, the degree of the left side is $1$ and the degree of your expression is $\frac{3}{2},$
which says that after homogenization we'll get a wrong inequality for $d\rightarrow0^+$.
Indeed, it's enough to prove that:
$$\frac{a+b+c+d}{4}\geq\sum_{cyc}\frac{\sqrt{abc}}{4}$$ or
$$(a+b+c+d)\sqrt[8]{abcd}\geq\sqrt{abc}+\sqrt{abd}+\sqrt{acd}+\sqrt{bcd}.$$
The last inequality is homogeneous already and we can forget about the condition $abcd=1$.
Now, let $d\rightarrow0^+$ and $a=b=c=1$.
Thus, the left side close to $0$ and the right side close to $1$, which says that this inequality is wrong.
The solution by Tangent Line method.
By AM-GM
$$\sum_{cyc}\frac{1}{a^3+b+c+d}\leq\sum_{cyc}\frac{1}{a^3+3\sqrt[3]{bcd}}=\sum_{cyc}\frac{1}{a^3+\frac{3}{\sqrt[3]a}}.$$
Now, let $a=x^3$, $b=y^3$, $c=z^3$ and $d=t^3$.
Thus, $xyzt=1$ and we need to prove that:
$$\sum_{cyc}\left(\frac{x^3}{4}-\frac{x}{x^{10}+3}\right)\geq0.$$
Now, since by AM-GM
$$\frac{1}{3}x^{10}-\frac{10}{3}x+3\geq0$$ and $$\frac{2}{3}x^{10}-2x^4+\frac{4}{3}x\geq0,$$ after summing we obtain: $$x^{10}+3\geq2x^4+2x$$ and it's enough to prove that
$$\sum_{cyc}\left(\frac{x^3}{4}-\frac{x}{2x^4+2x}\right)\geq0$$ or
$$\sum_{cyc}\left(x^3-\frac{2}{x^3+1}\right)\geq0$$ or
$$\sum_{cyc}\left(a-\frac{2}{a+1}-\frac{3}{2}\ln{a}\right)\geq0,$$ which is true because
$$\left(a-\frac{2}{a+1}-\frac{3}{2}\ln{a}\right)'=\frac{(a-1)(2a^2+3a+3)}{2a(a+1)^2}.$$
The coefficient $\frac{3}{2}$ we can get by the following way.
Let $f(x)=x-\frac{2}{x+1}+\lambda\ln{x}$.
We see that $f(1)=0$.
We'll choose $\lambda$ such that also $f'(1)=0$.
Easy to see that it gives $\lambda=-\frac{3}{2}.$
A: 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, $LHS<RHS$.
A: Another way.
By C-S $$\sum_{cyc}\frac{1}{a^3+b+c+d}=\sum_{cyc}\frac{\frac{1}{a}+b+c+d}{(a^3+b+c+d)\left(\frac{1}{a}+b+c+d\right)}\leq$$
$$\leq\frac{\sum\limits_{cyc}\left(\frac{1}{a}+b+c+d\right)}{(a+b+c+d)^2}=\frac{\sum\limits_{cyc}\left(abc+3a\right)}{(a+b+c+d)^2}.$$
Thus, it's enough to prove that:
$$(a+b+c+d)^3\geq4\sum_{cyc}abc+12(a+b+c+d).$$
But by Muirhead (or Maclurin) $$(a+b+c+d)^3\geq16\sum_{cyc}abc$$
and by AM-GM $$\frac{3}{4}(a+b+c)^3\geq\frac{3}{4}\left(4\sqrt[4]{abcd}\right)^2(a+b+c+d)=12(a+b+c+d)$$ and we are done!
