Let $a,b,c\in\mathbb{R}_{+}$, it's true that $$\sum_{cyc}{\frac{a}{3a+b}}\leq\frac{3}{4}$$
My approach: By AM-GM, $3a+b\geq 4(a^{3}b)^{1/4}$, then $\sum_{cyc}{\frac{a}{3a+b}}\leq\frac{1}{4}(\frac{a}{b})^{1/4}$, so I need prove that
$$\left(\frac{a}{b}\right)^{1/4}+\left(\frac{b}{c}\right)^{1/4}+\left(\frac{c}{a}\right)^{1/4}\leq3$$
Well, I need prove the above expression, when I tried to use AM-GM, I get
$$\left(\frac{a}{b}\right)^{1/4}+\left(\frac{b}{c}\right)^{1/4}+\left(\frac{c}{a}\right)^{1/4}\geq3\left[\left(\frac{a}{b}\right)^{1/4}\left(\frac{b}{a}\right)^{1/4}\left(\frac{c}{a}\right)^{1/4}\right]^{1/3}=3$$
We need to prove that $$\sum_{cyc}\left(\frac{a}{3a+b}-\frac{1}{3}\right)\leq\frac{3}{4}-1$$ or $$\sum_{cyc}\frac{b}{3a+b}\geq\frac{3}{4},$$ which is true by C-S: $$\sum_{cyc}\frac{b}{3a+b}=\sum_{cyc}\frac{b^2}{3ab+b^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3ab+b^2)}\geq\frac{3}{4},$$ where the last inequality it's just $$\sum_{cyc}(a-b)^2\geq0.$$