The inequality $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \frac{3}{2}$

Question

I was told that the following inequality

$$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \frac{3}{2}$$

can be solved by the rearrangement inequality https://en.wikipedia.org/wiki/Rearrangement_inequality

But I don't see how to do this. Is there some simple way to prove this inequality?

Answer 1

WLOG, let $x\le y\le z$. then $x+y\le x+z\le y+z$, so by rearrangement inequality, $$\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{x}{x+y} \\ \dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{z}{y+z}+\dfrac{x}{x+z}+\dfrac{y}{x+y}$$ Add the two inequalities, we get: $$2\left(\dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\right)\ge3 \\ \therefore \dfrac{x}{y+z}+\dfrac{y}{x+z}+\dfrac{z}{x+y}\ge\dfrac{3}{2}$$

Answer 2

Set $s=x+y+z$, then the left side is the cyclic sum over $\frac{x}{s-x}=\frac{s}{s-x}-1$. Considering $s$ fixed for the moment, this is a convex function for $0\le x<s$. By Jensen's inequality $$ \frac13\sum_{cyc}\frac{x}{s-x}=\frac13\sum_{cyc}f(x)\ge f\left(\frac13\sum_{cyc}x\right)=\frac{s/3}{s-s/3}=\frac12. $$

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One does not need the general Jensen inequality, one can also use the inequality of harmonic and arithmetic mean (which is Jensen's inequality for $f(x)=\frac1x$), $$ \frac{\frac{s}{s-x}+\frac{s}{s-y}+\frac{s}{s-z}}3\ge\frac3{\frac{s-x}{s}+\frac{s-y}{s}+\frac{s-z}{s}}=\frac32 $$ so that $$ \sum_{cyc}\frac{x}{s-x}\ge\frac92-3=\frac32 $$

Answer 3

For positive variables need to prove that $$\sum_{cyc}\left(\frac{x}{y+z}+1\right)\geq\frac{3}{2}+3$$ or $$\sum_{cyc}\frac{x+y+z}{y+z}\geq\frac{9}{2}$$ or $$\sum_{cyc}(y+z)\sum_{cyc}\frac{1}{y+z}\geq9.$$ Now, since $(x+y,x+z,y+z)$ and $\left(\frac{1}{x+y},\frac{1}{x+z},\frac{1}{y+z}\right)$ have an opposite ordering, by Rearrangement we obtain: $$\sum_{cyc}(y+z)\sum_{cyc}\frac{1}{y+z}\geq3\left((y+z)\cdot\frac{1}{y+z}+(x+z)\cdot\frac{1}{x+z}+(x+y)\cdot\frac{1}{x+y}\right)=9$$ and we are done!