Get the answer to "Let ABC be an acute triangle. What is..."

Destinee Bryan

Answered question

2022-04-16

Let ABC be an acute triangle. What is the maximum value of

\frac{\tan^{2} A + \tan^{2} B}{\tan^{4} A + \tan^{4} B} + \frac{\tan^{2} B + \tan^{2} C}{\tan^{4} B + \tan^{4} C} + \frac{\tan^{2} C + \tan^{2} A}{\tan^{4} C + \tan^{4} A} ?

For the equilateral triangle,

\tan A = \tan B = \tan C = \sqrt{3}

, and the sum is 1. For the isosceles triangle with

∠ A = ∠ B \to \frac{π}{2}

and

∠ C \to 0

, the sum approaches zero.

zvonkurm7h

Beginner2022-04-17Added 8 answers

Say

x = \tan A, y = \tan B and z = \tan C

. It is well-known that x+y+z=xyz. Then we have

E = \frac{x^{2} + y^{2}}{x^{4} + y^{4}} + \frac{y^{2} + z^{2}}{y^{4} + z^{4}} + \frac{z^{2} + x^{2}}{z^{4} + x^{4}}

By Cauchy inequality and then by AM-GM we have

\frac{x^{2} + y^{2}}{x^{4} + y^{4}} \leq \frac{2}{x^{2} + y^{2}} \leq \frac{1}{x y}

E \leq \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x} = 1

We get equality exactly when x=y=z, thus

3 x = x^{3} \to x = \sqrt{3}

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