an inequality

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High School Math Grades 9-12, Ages 13-18

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#1 h Tuesday at 7:56 AM

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Let a,b,c\ge 0: ab+bc+ca>0 then prove \frac{ab+c^2}{a+b}+\frac{bc+a^2}{b+c}+\frac{ca+b^2}{c+a}\ge\frac{2(a^2+b^2+c^2)+ab+bc+ca}{a+b+c}.

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#2 h Tuesday at 10:18 AM

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WLOG that $a = \min\{a, b, c\}.$ Note that
$\frac{2\sum a^2 + \sum bc}{\sum a} = \sum a + \frac{\sum a^2 - \sum bc}{\sum a} \le \sum a + \frac{\sum a^2 - \sum bc}{b+c} = \frac{a^2 + 2b^2 + 2c^2 + bc}{b+c}.$ Hence, it suffices to prove that
$\sum \frac{c^2 + ab}{a+b} \ge \frac{a^2 + 2b^2 + 2c^2 + bc}{b+c},$ the term $\frac{a^2+bc}{b+c}$ cancels on both sides, and is hence equivalent to,
$\frac{c^2 + ab}{a+b} + \frac{b^2 + ac}{a+c} \ge \frac{2(b^2 + c^2)}{b+c}.$ which is equivalent to
$(b-c)^2 \Bigl[(b+c)^2 - (a+b)(a+c)\Bigr] \ge 0,$ which is evidently true. Therefore, the original inequality holds.

This post has been edited 2 times. Last edited by lbh_qys, Tuesday at 10:34 AM

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