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Let x, y, z be positive real numbers. Prove that $$\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y} \ge \frac{3}{2}$$

This problem appears to be simple, but upon further work and lots of failed attempts, I am stuck. I have tried using arithmetic and harmonic means (which I am sure are the key) to show that there exists some number which is fits between these two, thus proving the inequality. I have also tried multiplying it out and simplifying and obtained:

$$1 + \frac{x^3 + y^3 + z^3 + xyz}{(x+y)(y+z)(z+x)} \ge \frac{3}{2} $$ This didn't really seem to help. Any guidance is greatly appreciated!

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$$A=\dfrac a{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}$$ $$B= \dfrac b{b+c}+ \dfrac{c}{c+a}+\dfrac{a}{a+b}$$ $$C=\dfrac c{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+a}$$

By A-G inequality

$$A+B=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+c}+\dfrac{c+a}{b+a}\ge 3$$ $$A+C=\dfrac{a+c}{b+c}+\dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}\ge3$$

so + $$(A+B)+(A+C)\ge6$$

$B+C=3$, we get $2A\ge3$, yeah

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Let $s = x+y+z$. The function $$f:t \mapsto \frac{t}{s-t}$$ for $t \in [0, s)$ is convex. Therefore $$\frac{f(x) + f(y) + f(z)}{3} \geq f\left( \frac{s}{3}\right)$$ and the latter is exactly your inequality.

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  • $\begingroup$ That's a really beautiful argument but not how I was hoping to show the result! $\endgroup$ – Georgia Oct 17 '13 at 19:46
  • $\begingroup$ @Georgia The links in Macavity's comment seem to provide ample alternatives. $\endgroup$ – WimC Oct 17 '13 at 19:47
  • $\begingroup$ Indeed, but I always appreciate having more than one approach for future reference. Thanks for the great (and elegant) response. $\endgroup$ – Georgia Oct 17 '13 at 19:49

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