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Difference between revisions of "1980 USAMO Problems/Problem 5"
 
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== See Also ==
 
== See Also ==
 
{{USAMO box|year=1980|num-b=4|after=Last Question}}
 
{{USAMO box|year=1980|num-b=4|after=Last Question}}
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[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]

Latest revision as of 19:11, 3 July 2013

Problem

If $x, y, z$ are reals such that $0\le x, y, z \le 1$, show that $\frac{x}{y + z + 1} + \frac{y}{z + x + 1} + \frac{z}{x + y + 1} \le 1 - (1 - x)(1 - y)(1 - z)$

Solution

Rewrite the given inequality so that $1$ is isolated on the right side. Set the left side to be $f(x, y, z)$. Now a routine computation shows


$\frac{\partial^2 f}{\partial x^2} = \frac{2y}{(x + z + 1)^3} + \frac{2z}{(x + y + 1)^3}\geq 0$


which shows that $f$ is convex (concave up) in all three variables. Thus the maxima can only occur at the endpoints, i.e. if and only if $x, y, z \in \{0,1\}$. Checking all eight cases shows that the value of the expression cannot exceed 1.

See Also

1980 USAMO (Problems • Resources)
Preceded by
Problem 4 		Followed by
Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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