An algebraic inequality in three real variables Is it true that
$$(v-u)^2+(w-u)^2+(w-v)^2 \\ 
+\left(\sqrt{\frac{1+u^2}{1+v^2}}
+\sqrt{\frac{1+v^2}{1+u^2}}\right) (w-u)(w-v) \\ 
-\left(\sqrt{\frac{1+u^2}{1+w^2}}+\sqrt{\frac{1+w^2}{1+u^2}}\right) (w-v)(v-u) \\ 
-\left(\sqrt{\frac{1+w^2}{1+v^2}}+\sqrt{\frac{1+v^2}{1+w^2}}\right) (v-u) (w-u)>0$$
for any real $u,v,w$ such that $u<0<v<w$?
Certain numerical evidence suggests it is true.
This inequality arose in the previous answer.

Rewrite the inequality in terms of $a,b,c>1$ via $u=(\frac{1}{a}-a)/2$, $v=(b-\frac{1}{b})/2$, and $w=(c-\frac{1}{c})/2$. Note that $\sqrt{1+u^2}=\frac{1+a^2}{2a}$ and so on. Then the conjectured inequality is equivalent to
\begin{equation}
 ((c - b)(a + c)(a + b))^2(a^2b^2c^2-a^2bc+ab^2c+abc^2+ab+ac-bc+1)>0.
\end{equation}
Note that the second factor equals
\begin{equation}
a^2bc(bc-1) + abc(b+c-1) + ab + ac +1,
\end{equation}
which is positive since $b,c>1$.
