Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$

Question

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

Answer 1

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.

Answer 2

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!