Wednesday, December 2, 2009

Solution: Inequality

Original problem:

For positive real number $a,b,c$ prove that

$2\sqrt{ab+bc+ca} \leq \sqrt{3} \sqrt[3]{(a+b)(b+c)(c+a)}$

First Solution

Credit for this solution goes to Sander Parawira

$!9 (a+b)(a+c)(b+c) \geq 8 (a+b+c)(ab+ac+bc) \geq 8 \sqrt{3} (ab+ac+bc)^{3/2}$

The first inequality is well-known, and the second one is equivalent to $a^2+b^2+c^2 \geq ab+bc+ca$

$! 3 \sqrt{3} (a+b)(a+c)(b+c) \geq 8 (ab+ac+bc)^{3/2}$
$! \sqrt{3} \sqrt[3]{ (a+b)(a+c)(b+c) } \geq 2 \sqrt{ab+ac+bc}$

Second Solution

Since it's homogeneous, we may assume that $ab+bc+ca=1$. We wish to minimize $(a+b)(b+c)(c+a)$.

If we let $a = \cot A, b = \cot B, c = \cot C$, then the normalization condition implies that $A,B,C$ are angles of a triangle.

$(a+b)(a+c) = a^2+ab+bc+ca = a^2+1 = \csc^2 A$

So $(a+b)(b+c)(c+a) = \csc A \csc B \csc C = 2R^2/S$ where $S$ is the area of the triangle and $R$ is the circum-radius.

Now the problem can be rewritten as: given a circle with radius 1, find 3 points $ABC$ on the circle such that the triangle $ABC$ has maximal area.

We claim that the maximal area occurs when $ABC$ is an equilateral triangle. Suppose $AB \neq BC$, then let $B'$ the midpoint of the arc $AC$ that passes through $B$. Compare the area of $ABC$ and $AB'C$. These triangles have the same base $AC$, but different heights. Since $B'$ is the midpoint of the arc, then the distance from $B'$ to $AC$ is greater than that of $B$ to $AC$. Thus $S_{ABC} < S_{AB'C}$.

This means that whenever the triangle has two unequal sides, we can always find another triangle with greater area. The process stops when all three sides are equal, and that's when $ABC$ is an equilateral triangle.

Variation: minimizing $\csc A \csc B \csc C$ can also be done with Jensen, using the function $f(x) = \log \csc x$

This is because $f'(x) = -\cot(x)$ and $f''(x) = \csc^2 x$ thus $f$ is convex.

1 comment:

  1. [...] (1) Comment Read [...]