Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$ Let $ABC$ be acute triangle. Prove that
$$\sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$$
My attempt:
$$\sqrt{\tan A} + \sqrt{\tan B}\geq 2\sqrt[4]{\tan A\cdot\tan B}$$
At here I think I need to prove $$2\sqrt[4]{\tan A\cdot\tan B}\ge2\sqrt{\cot\frac{A}{2}}$$
$$\Leftrightarrow \tan A\cdot\tan B\ge \cot^2\frac{A}{2}$$
And I was stuck here.
Help me... Thanks
 A: Remark: $\tan A,\ \tan B,\ \tan C,\ \cot\frac{A}{2},\ \cot\frac{B}{2},$ and $\cot\frac{C}{2}$ are all positive because $0<A,\ B,\ C<\frac{\pi}{2}$. This is used in this proof, for example in cancellings or taking square roots of both sides of inequalities.
The function $f\left(x\right)=\tan x$ is convex on the interval $\left(0,\ \frac{\pi}{2}\right)$, so we can use Jensen's Inequality to show that
$\begin{align}
\sum_\text{cyc}\frac{\tan A+\tan B}{2}\ge\sum_\text{cyc}\tan\frac{A+B}{2}=\sum_\text{cyc}\cot\frac{C}{2}\Rightarrow\sum_\text{cyc}\tan A\ge\sum_\text{cyc}\cot\frac{A}{2}\ \left(\#\right)
\end{align}$
It remains to prove that $\sum_\limits\text{cyc}\sqrt{\tan A\tan B}\ge\sum_\limits\text{cyc}\sqrt{\cot\frac{A}{2}\cot\frac{B}{2}}\ \left(\#\#\right)\Leftrightarrow\\$
$\begin{align}
&\Leftrightarrow\sum_\text{cyc}\tan A\tan B\ge\sum_\text{cyc}\cot\frac{A}{2}\cot\frac{B}{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\frac{2\cot\frac{A}{2}}{\cot^{2}\ \frac{A}{2}-1}\cdot\frac{2\cot\frac{B}{2}}{\cot^{2}\ \frac{B}{2}-1}\ge\sum_\text{cyc}\cot\frac{A}{2}\cot\frac{B}{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\frac{4}{\left(\cot^{2}\frac{A}{2}-1\right)\left(\cot^{2}\frac{B}{2}-1\right)}\ge0\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\left(\cot^{2}\frac{A}{2}-1\right)\left(\cot^{2}\frac{B}{2}-1\right)\ge0
\end{align}$
which is true because $0<A,\ B<\frac{\pi}{2}\Leftrightarrow$
$\Leftrightarrow1<\cot\frac{A}{2},\ \cot\frac{A}{2}<+\infty$.
From $\left(\#\right)$ and $\left(\#\#\right)$ we have
$\begin{align} 
&\sum_\text{cyc}\tan A+2\sum_\text{cyc}\sqrt{\tan A\tan B}\ge\sum_\text{cyc}\cot\frac{A}{2}+2\sum_\text{cyc}\sqrt{\cot\frac{A}{2}\cot\frac{B}{2}}\Leftrightarrow\\
&\Leftrightarrow\left(\sum_\text{cyc}\sqrt{\tan A}\right)^{2}\ge\left(\sum_\text{cyc}\sqrt{\cot\frac{A}{2}}\right)^{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\sqrt{\tan A}\ge\sum_\text{cyc}\sqrt{\cot\frac{A}{2}}\ \cdot
\end{align}$
Done!
A: This is not an answer, but provides a geometric understanding of the desired inequality.
Let $A,B,C$ be the three angles of the acute triangle, and then let $$A'=\frac{\pi}{2}-\frac{A}{2},\, B'=\frac{\pi}{2}-\frac{B}{2},\, C'=\frac{\pi}{2}-\frac{C}{2}$$
Hence $A', B', C'$ are angles of a new acute triangle. The inequality says that
$$\sqrt{\tan A}+\sqrt{\tan B}+\sqrt{\tan C}\ge \sqrt{\tan A'}+\sqrt{\tan B'}+\sqrt{\tan C'}. $$
The picture can be visualized as follows:
Let $ABC$ be the orginal triangle with vertices $A,B,C$ corresponding to the angles $A,B,C$. (There is no ambiguation here, I think.) Let $l_A,l_b$ be bisector lines starting from $A$ and $B$, and let $l_A',l_B'$ be the lines perpendicular to $l_A, l_B$ and passing through $A$ and $B$ respectively, and thus they intersect at a point $D$. Then the triangle $ABD$ is the triangle $A'B'C'$ constructed above, with $A=A', B=B', D=C'$.
I think the inequality is rather interesting. I'm sure that it can be proved by calculus, but a fundamental geometric proof would be more favourable.
