Let $a,b,c\ge 0: ab+bc+ca=3$. Prove that$$\frac{1}{\sqrt{5a+4}}+\frac{1}{\sqrt{5b+4}}+\frac{1}{\sqrt{5c+4}}\geq1.$$
This problem is from an AOPS account- arqady. Sorry, I can not find the link.
The equality is $a=b=c=1$, which implies me using $AM-GM:$
$$\sum_{cyc}\frac{1}{\sqrt{5a+4}}\ge 6\sum_{cyc}\frac{1}{5a+13}$$And we need to prove$$\frac{1}{5a+13}+\frac{1}{5b+13}+\frac{1}{5c+13}\ge \frac{1}{6}$$But it is wrong at $a=b=\dfrac{1}{5}$.
I hope to see some ideas. Thank you very much.
Best Answer
By AM-GM, we have $$\sum_{\mathrm{cyc}} \frac{1}{\sqrt{5a+4}} = \sum_{\mathrm{cyc}} \frac{2}{2\sqrt{\frac{5a+4}{a + 2}\cdot (a + 2)}} \ge \sum_{\mathrm{cyc}} \frac{2}{\frac{5a + 4}{a + 2} + a + 2}.$$
It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{2}{\frac{5a + 4}{a + 2} + a + 2} \ge 1. \tag{1}$$
We use pqr method.
Let $p = a + b + c, q = ab + bc + ca = 3, r = abc$.
(1) is written as $$-r^2 - (37p + 204)r + 68p + 112 \ge 0. \tag{2}$$
Using $q^2 \ge 3pr$, it suffices to prove that $$- \left(\frac{3}{p}\right)^2 - (37p + 204)\cdot \frac{3}{p} + 68p + 112 \ge 0$$ or $$\frac{(p - 3)(68p + 1)(p + 3)}{p^2} \ge 0$$ which is true using $p^2 \ge 3q$.
We are done.