ant_ wrote: If $a,b,c\in(0,\infty)$ such that $a+b+c=3$, then $$\frac{a}{3a+bc+12}+\frac{b}{3b+ca+12}+\frac{c}{3c+ab+12}\le \frac{3}{16}$$ Proof of Zhangyanzong:

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sqing wrote: ant_ wrote: If $a,b,c\in(0,\infty)$ such that $a+b+c=3$, then $$\frac{a}{3a+bc+12}+\frac{b}{3b+ca+12}+\frac{c}{3c+ab+12}\le \frac{3}{16}$$ Proof of Zhangyanzong: Wow amazing!

By Cauchy-Schwarz Inequality, we have $$\frac{16}{3a+bc+12}\le \frac{1}{3a+bc}+\frac{9}{12}=\frac{1}{3a+bc}+\frac{3}{4}$$so $$\sum_{cyc}\frac{a}{3a+bc+12}\le \frac{1}{16}\left(\sum_{cyc}\frac{a}{3a+bc}+\frac94\right).$$But \begin{align*} \sum_{cyc}\frac{a}{3a+bc}=&\sum_{cyc}\frac{a}{(a+b)(a+c)}=\frac{2(ab+bc+ca)}{(a+b)(b+c)(c+a)}\\ =&\frac{2(a+b+c)(ab+bc+ca)}{3(a+b)(b+c)(c+a)}\le\frac34

\end{align*}so, see here,

done.

ant_ wrote: If $a,b,c\in(0,\infty)$ such that $a+b+c=3$, then $$\frac{a}{3a+bc+12}+\frac{b}{3b+ca+12}+\frac{c}{3c+ab+12}\le \frac{3}{16}$$ By AM-GM, $$\sum_{cyc}\frac{a}{3a+bc+12}=\sum_{cyc}\frac{a}{(c+a)(a+b)+4+4+4}\leq\frac{1}{4}\sum_{cyc}\frac{a}{\sqrt[4]{64(c+a)(a+b)}}=\frac{1}{16}\sum_{cyc}\sqrt[4]{\frac{2a}{c+a}\cdot\frac{2a}{a+b}\cdot a\cdot a}\leq \frac{1}{64}\sum_{cyc}\left(\frac{2a}{c+a}+\frac{2a}{a+b}+ a+ a\right)= \frac{3}{16}$$