Difference between revisions of "1991 APMO Problems/Problem 3"
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| − | By Cauchy-Schwarz, <math>\left(\sum \dfrac{a_i^2}{a_i+b_i}\right)\left(\sum a_i+b_i\right)\geq \left(\sum a_i\right) ^2</math>, so <math>\sum \dfrac{a_i^2}{a_i+b_i}\geq \dfrac{\left(\sum a_i\right) ^2}{\sum (a_i+b_i)}=\dfrac{\sum a_i}{2} since < | + | By Cauchy-Schwarz, <math>\left(\sum \dfrac{a_i^2}{a_i+b_i}\right)\left(\sum a_i+b_i\right)\geq \left(\sum a_i\right) ^2</math>, so <math>\sum \dfrac{a_i^2}{a_i+b_i}\geq \dfrac{\left(\sum a_i\right) ^2}{\sum (a_i+b_i)}=\dfrac{\sum a_i}{2}</math> since <math>\sum a_i=\sum b_i</math>. |
Revision as of 16:40, 4 August 2011
Problem
Let 
, 
, 
, 
, 
, 
, , 
be positive real numbers such that 
. Show that
![\[\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2}\]](http://latex.artofproblemsolving.com/0/f/4/0f4985029cc0227d46a829d94c18ab1432ac8dc6.png)
Solution
By Cauchy-Schwarz, 
, so 
since 
.
