How to prove elegantly this inequality? Let $a$, $b$ and $c$ three reals greater than or equal to 1. Put $a+b+c=3u$, $ab+bc+ca=3v^2$ and $abc=w^3$. Show that
$$w-\frac{1}{w}\ge\frac{w^3}{v^2}-\frac{1}{u}$$
I am able to prove it by assuming without loss of generality that $c\ge b\ge a$ and posing $b=a+u$ and $c=a+u+v$ for $u,v\ge0$. The problem here is that when plug  this two expression and develop, I get a huge polynomial of $a$ whose coefficients are all positive. I am wondering if there a possibility to obtain a more elegant solution.
 A: Write the question as
$$
\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \ge\!\!? \; \sqrt[3]{abc}  \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} - \frac{1}{\sqrt[3]{abc}} \frac{3 \sqrt[3]{abc}}{a+b+c}
$$
By AM-GM, both following coefficients are at most $1$:
$$
A =  \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} \le  1\qquad 
B = \frac{3 \sqrt[3]{abc}}{a+b+c}\le 1$$
Again, rewrite the question as
$$
\sqrt[3]{abc} - \frac{1}{\sqrt[3]{abc}} \ge\!\!? \; \frac{B+A}{2}\Big(\sqrt[3]{abc}  - \frac{1}{\sqrt[3]{abc}} \Big) - \frac{B-A}{2}\Big(\sqrt[3]{abc}  + \frac{1}{\sqrt[3]{abc}} \Big) 
$$
So the inequality is immediately proved for the case that
$$
 B =   \frac{3 \sqrt[3]{abc}}{a+b+c} \ge \frac{3\sqrt[3]{a^2b^2c^2}}{ab+bc+ca} = A\\
\leftrightarrow ab + bc + ca \ge (a+b+c)\sqrt[3]{abc}
$$
Now this is not generally true. By symmetry, let w.l.o.g. $c \ge b \ge a$ and realize this by writing $b=a x$ and $c = a x^2 q^3$. Since, by the original question, $a,b,c \ge 1$, we have  $a \ge 1$ and $ x\ge 1$ and $ q^3 x\ge 1$ .  Then we get the equivalent
$$
1 +  x q^3 + x^2 q^3- (1 + x +  x^2 q^3)q \ge 0 \\
\leftrightarrow q^3(1-q)(x-1/q)(x-1/q^2) \ge 0
$$
which is true (since $ x\ge 1$) for
$q \le 1$ and either $x<1/q$ or $x >1/q^2$.
So we need to consider the remaining cases 1.) $q > 1$, and  2.) $q < 1$ and $1/q<x <1/q^2$.
Taking a closer look at case 2.) we can write the condition as $q^2< q^3 x < q$ but,  since also $q <1$ in this case, this never meets the required  $ q^3 x\ge 1$. So case 2.) needs not be considered.
We have case 1.) $q > 1$: Using our substitutions, the original inequality gets
$$
a^2x^2q^2\Big( 1 -  \frac{3xq^2}{1 +  x q^3 + x^2 q^3}  \Big) \ge\!\!? \; 1 -  \frac{3xq}{1 +  x + x^2 q^3} 
$$
Again with $a \ge 1$ and $ x\ge 1$, we can prove the stronger inequality
$$
q^2 -  \frac{3xq^4}{1 +  x q^3 + x^2 q^3}   \ge\!\!? \; 1 -  \frac{3xq}{1 +  x + x^2 q^3} 
$$
Consider first $q$ fixed and $x$ variable. If $x$ increases, the inequality gets less tight. Inspecting  the derivatives w.r.t. $x$, the difference between the LHS and the RHS increases for $x > q^{-3/2}$ and as we have $q>1$ and $x\ge1$, this is always the case. So we only need to consider the tightest case for $x=1$. This leaves us to show
$$
q^2\Big( 1 -  \frac{3q^2}{1 +  2 q^3 }  \Big)  - \Big( 1 -  \frac{3q}{2 +   q^3} \Big)\ge\!\!? \; 0
$$
or
$$
\frac{(q-1)^3(2q^5 + 3q^4 + q^3 + q^2 + 3q + 2)}{(1 +  2 q^3) (2 +   q^3)} \ge\!\!? \; 0 
$$
which is true, since we consider $q>1$.
$\qquad \Box$
