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Trigonometric substitution

Understand the problem

Let $0 < a, b, c < 1$ with $ab + bc + ca = 1$. Prove that
\[\frac{a}{1-a^2} + \frac{b}{1-b^2} + \frac{c}{1-c^2} \geq \frac {3 \sqrt{3}}{2}.\]

Determine when equality holds.

Source of the problem
Singapore Team Selection Test 2004
Topic
Inequalities
Difficulty Level
Medium
Suggested Book
Inequalities by BJ Venkatachala

Start with hints

Do you really need a hint? Try it first!

Show that there exists a triangle \Delta ABC such that a=\tan\frac{A}{2}, b=\tan\frac{B}{2} and c=\tan\frac{C}{2}. The easiest way to prove is is to define A:=2\arctan a, B:=2\arctan b and show that \tan\frac{\pi-A-B}{2} must be c.

Now the inequality becomes equivalent to \tan A+\tan B+\tan C\ge 3\sqrt{3}. Take a look at this well-known inequality. Note that the tangent is a convex function in (0,\pi/2).

Note that, a,b,c are less than 1. This means that \frac{A}{2},\frac{B}{2} and \frac{C}{2} are all less than \frac{\pi}{4}. Hence the triangle is acute.

Using Jensen’s inequality, we get \frac{\tan A+\tan B+\tan C}{3}\ge\tan\frac{A+B+C}{3}=\tan\frac{\pi}{3}=\sqrt{3}. Equality holds for an equilateral triangle.

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