Difference between revisions of "2006 IMO Shortlist Problems/A5"

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[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Latest revision as of 08:03, 29 March 2008

Problem

(Hojoo Lee, South Korea) Let $a,b,c$ be the sides of a triangle. Prove that \[\frac{\sqrt{b+c-a}}{\sqrt{b}+ \sqrt{c} - \sqrt{a}} + \frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a+b-c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} \le 3.\]

This problem also appeared on the 2007 IMO TSTs of Italy and Bangladesh.

Solution

Lemma. For any positive reals $x,y,z$, \[\sum_{\rm cyc} \left( \frac{y}{x} - 1 \right) \left( \frac{z}{x} -1 \right) \ge 0 .\]

Proof 1. This is the Vornicu-Schur Inequality on the function $x \mapsto 1/x$. $\blacksquare$

Proof 2. Without loss of generality, suppose that $x\ge y \ge z$. Evidently, \[\left( \frac{y}{x} - 1 \right) \left( \frac{z}{x} - 1\right) \ge 0 ,\] so it suffices to show that \[\left( \frac{x}{z} - 1 \right) \left( \frac{y}{z} - 1 \right) + \left( \frac{x}{y} - 1 \right) \left( \frac{z}{y} - 1 \right) \ge 0,\] or \[\left( \frac{x}{z} - 1 \right) \left( \frac{y}{z} - 1 \right) \ge \left( \frac{x}{y} - 1 \right) \left( 1 - \frac{z}{y} \right) .\] By assumption, \[\frac{x}{z} - 1 \ge \frac{x}{y} - 1 ,\] and by AM-GM, $y/z + z/y \ge 2$, or \[\frac{y}{z} - 1 \ge 1 - \frac{z}{y} .\] Since all sides of both the previous two inequalities are positive, multiplication yields the desired bound, proving the lemma. $\blacksquare$

We note that in the problem statement, each denominator is greater than zero, for \[\sqrt{b} + \sqrt{c} = \sqrt{b + c + 2 \sqrt{bc}} \ge \sqrt{a + 2\sqrt{bc}} > \sqrt{a} .\] We now abbreviate $x = \sqrt{b} + \sqrt{c} - \sqrt{a}$, $y = \sqrt{c} + \sqrt{a} - \sqrt{b}$, $z= \sqrt{a} + \sqrt{b} - \sqrt{c}$. Then \begin{align*} \frac{\sqrt{b+c-a}}{\sqrt{b}+ \sqrt{c} - \sqrt{a}} + \frac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a+b-c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} &= \sum_{\rm cyc} \sqrt{ \frac{ \left( \frac{x+y}{2} \right)^2 + \left( \frac{x+z}{2} \right)^2 - \left( \frac{y+z}{2} \right)^2 }{x^2}} \\ &= \frac{1}{2} \sum_{\rm cyc} \sqrt{ \frac{2x^2 + 2xy + 2xz - 2yz}{x^2}} .  \end{align*} By the Power Mean Inequality, \[\frac{1}{2} \sum_{\rm cyc} \sqrt{ \frac{2x^2 + 2xy + 2xz - 2yz}{x^2}} \le \frac{\sqrt{3}}{2} \biggl( \sum_{\rm cyc} \frac{2x^2 + 2xy + 2xz - 2yz}{x^2} \biggr)^{1/2} .\] Also, by the lemma, \[0 \le \sum_{\rm cyc} \left(\frac{y}{x}-1 \right) \left( \frac{z}{x} -1 \right) = \sum_{\rm cyc} \frac{yz - xz - xy}{x^2} + 1 = 6 - \sum_{\rm cyc} \frac{x^2 + xy + xz - yz}{x^2} ,\] so \[\frac{\sqrt{3}}{2} \biggl( \sum_{\rm cyc} \frac{2x^2 + 2xy + 2xz - 2yz}{x^2} \biggr)^{1/2} \le \frac{\sqrt{3}}{2} ( 2 \cdot 6 )^{1/2} = 3,\] proving the desired inequality. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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