Prove $\frac{bc}{\sqrt{7a+9}}+\frac{ca}{\sqrt{7b+9}}+\frac{ab}{\sqrt{7c+9}}\le \frac{3}{4}; a+b+c=3.$

Question

Given non-negative real numbers $a,b,c$ satisfying $a+b+c=3.$ Prove that $$\frac{bc}{\sqrt{7a+9}}+\frac{ca}{\sqrt{7b+9}}+\frac{ab}{\sqrt{7c+9}}\le \frac{3}{4}.$$ Equality holds at $a=b=c=1$ or $a=b=\dfrac{3}{2};c=0.$ By using Cauchy-Schwarz inequality $$\sum_{cyc}\frac{bc}{\sqrt{7a+9}}=\sum_{cyc}\sqrt{bc}\cdot\sqrt{\frac{bc}{7a+9}}\le \sqrt{(ab+bc+ca)\cdot\sum_{cyc}\frac{bc}{7a+9}}.$$ And we need to prove $$\color{black}{\frac{bc}{7a+9}+\frac{ca}{7b+9}+\frac{ab}{7c+9}\le \frac{9}{16(ab+bc+ca)}. }$$ I think the last one is good because it saves the case of equality occuring.

I tried to homogenize but it's complicated. Hope you help me continue my approach.

Answer 1

Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that{ $$\sum_{cyc}\frac{bc}{7a+9u}\leq\frac{3u^3}{16v^2},$$ which is a linear inequality of $w^3$, which by $uvw$ say that it's enough to check two cases:

1. $c=0$ and $a+b=3$.

Thus, we need to prove that: $$\frac{ab}{9}\leq\frac{9}{16ab},$$ which is true by AM-GM;

2. Two variables are equal.

Let $b=a$ and $c=3-2a$, where $0\leq a\leq\frac{3}{2}.$

Thus, we need to prove that: $$(3-2a)(a-1)^2(45+134a-84a^2)\geq0$$ and we are done.