Proving Sum of Tangents $\ge$ Sum of Cotangents in Acute Triangle

[anemone]

In an acute triangle $ABC$, prove that $\sqrt{\tan A}+\sqrt{\tan B}+\sqrt{\tan C} \ge \sqrt{\cot \left(\dfrac{A}{2}\right)}+\sqrt{\cot \left(\dfrac{B}{2}\right)}+\sqrt{\cot \left(\dfrac{C}{2}\right)}$.

 

[jvicens]

Thank you for your post regarding the inequality in an acute triangle $ABC$: $\sqrt{\tan A}+\sqrt{\tan B}+\sqrt{\tan C} \ge \sqrt{\cot \left(\dfrac{A}{2}\right)}+\sqrt{\cot \left(\dfrac{B}{2}\right)}+\sqrt{\cot \left(\dfrac{C}{2}\right)}$.

First, let us recall some basic definitions and properties of trigonometric functions in an acute triangle $ABC$. In an acute triangle, all angles $A$, $B$, and $C$ are less than $90$ degrees, and therefore, all trigonometric functions are positive. Additionally, we know that the sum of the angles in a triangle is equal to $180$ degrees, which means that $A+B+C=180$.

Now, let us consider the left side of the inequality, $\sqrt{\tan A}+\sqrt{\tan B}+\sqrt{\tan C}$. We can rewrite this as $\sqrt{\tan A}+\sqrt{\tan B}+\sqrt{\tan (180-A-B)}$, since $C=180-A-B$ in an acute triangle. Using the fact that $\tan (180-\theta)=\cot \theta$, we can simplify this to $\sqrt{\tan A}+\sqrt{\tan B}+\sqrt{\cot (A+B)}$.

Next, let us consider the right side of the inequality, $\sqrt{\cot \left(\dfrac{A}{2}\right)}+\sqrt{\cot \left(\dfrac{B}{2}\right)}+\sqrt{\cot \left(\dfrac{C}{2}\right)}$. Using the double angle formula for cotangent, we can rewrite this as $\sqrt{\frac{1+\cos A}{1-\cos A}}+\sqrt{\frac{1+\cos B}{1-\cos B}}+\sqrt{\frac{1+\cos C}{1-\cos C}}$.

Now, we can use the fact that $1-\cos A=\sin^2 A$ and $1+\cos A=2\cos^2 \frac{A}{2}$ to further simplify the right side to $2\sqrt{\frac{\cos^2 \frac{A}{2}}{\sin^2 A}}+2\sqrt{\frac{\cos^2 \frac{B}{2