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Difference between revisions of "1991 APMO Problems/Problem 3"

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== Solution ==
 
== Solution ==
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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}</math> since <math>\sum a_i=\sum b_i</math>.
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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>. Since <math>\sum a_i=\sum b_i</math>, we get<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>.

Latest revision as of 16:41, 4 August 2011

Problem

Let $a_1$, $a_2$, $\cdots$, $a_n$, $b_1$, $b_2$, $\cdots$, $b_n$ be positive real numbers such that $a_1 + a_2 + \cdots + a_n = b_1 + b_2 + \cdots + b_n$. Show that \[\frac{a_1^2}{a_1 + b_1} + \frac{a_2^2}{a_2 + b_2} + \cdots + \frac{a_n^2}{a_n + b_n} \geq \frac{a_1 + a_2 + \cdots + a_n}{2}\]

Solution

By Cauchy-Schwarz, $\left(\sum \dfrac{a_i^2}{a_i+b_i}\right)\left(\sum a_i+b_i\right)\geq \left(\sum a_i\right) ^2$. Since $\sum a_i=\sum b_i$, we get$\sum \dfrac{a_i^2}{a_i+b_i}\geq \dfrac{\left(\sum a_i\right) ^2}{\sum (a_i+b_i)}=\dfrac{\sum a_i}{2}$.

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