Given $a$, $b$, and $c$ are three positive real numbers. How to prove that sum of $\frac{a}{b+c}$, $\frac{b}{c+a}$ and $\frac{c}{a+b}$ is greater than or equal to $1.5$?
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3Here are six proofs. – Sahiba Arora Jun 19 '17 at 18:45
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by Cauchy Schwarz we have $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\geq \frac{(a+b+c)^2}{2(ab+bc+ca)}\geq \frac{3}{2}$$ if and only if $$a^2+b^2+c^2\geq ab+bc+ca$$ which is true.
Dr. Sonnhard Graubner
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