Difference between revisions of "1980 USAMO Problems/Problem 5"

(Created page with "== Problem == If <math>x, y, z</math> are reals such that <math>0\le x, y, z \le 1</math>, show that <math>\frac{x}{y + z + 1} + \frac{y}{z + x + 1} + \frac{z}{x + y + 1} \le 1 ...")
 
 
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== Solution ==
 
== Solution ==
{{solution}}
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Rewrite the given inequality so that <math>1</math> is isolated on the right side. Set the left side to be <math>f(x, y, z)</math>. Now a routine computation shows
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<math>\frac{\partial^2 f}{\partial x^2} = \frac{2y}{(x + z + 1)^3} + \frac{2z}{(x + y + 1)^3}\geq 0 </math>
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which shows that <math>f</math> is convex (concave up) in all three variables. Thus the maxima can only occur at the endpoints, i.e. if and only if <math>x, y, z \in \{0,1\}</math>. Checking all eight cases shows that the value of the expression cannot exceed 1.
  
 
== 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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{{MAA Notice}}
  
 
[[Category:Olympiad Inequality Problems]]
 
[[Category:Olympiad Inequality Problems]]

Latest revision as of 18:11, 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 (ProblemsResources)
Preceded by
Problem 4
Followed by
Last Question
1 2 3 4 5
All USAMO Problems and Solutions

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