Alternate proofs of the inequality $\sum_{\text{cyclic}} \frac{x}{2x+y+z}\le \frac{3}{4}$.

Question

If $x,y,z\in(0,\infty),$ prove the inequality, $$\sum_{\text{cyclic}} \frac{x}{2x+y+z}\le \frac{3}{4}$$ I have a solution using the substitution $x+y=a,$ $y+z=b$ and $z+x=c$. $$\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}=\frac{1}{2}\left(\frac{c+a-b}{c+a}+\frac{a+b-c}{a+b}+\frac{b+c-a}{b+c}\right)=\frac{1}{2}\left(3-\sum\frac{a}{b+c}\right)$$ Using Nesbitt's Inequality, $$\sum\frac{a}{b+c}\ge \frac{3}{2}\implies \frac{1}{2}\left(3-\sum\frac{a}{b+c}\right)\le \frac{3}{4}$$

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Is there a simpler solution to this problem that I am completely oblivious to? This was part of a problem set with relatively easier (well known) problems and I wouldn't be allowed to 'state' the Nesbitt's Inequality without proof.

Answer 1

Alternate proof:

$$\sum_{\text{cyclic}} \frac{x}{2x+y+z}\le \frac{3}{4}\iff\sum_{\text{cyclic}} \frac{x+y+z}{2x+y+z}\ge \frac{9}{4}\\\iff 4(x+y+z)\sum_{\text{cyclic}} \frac{1}{2x+y+z}\ge 9\\\iff\sum_{\text{cyclic}}2x+y+z \sum_{\text{cyclic}} \frac{1}{2x+y+z}\ge 9$$

This is obvious by Cauchy-Schwartz, or AM-GM, etc.

Answer 2

Just another way: set $x+y+z=3$, then use Jensen's inequality with the concave function $t \mapsto \dfrac{t}{t+3}$…