Let $a$,$b$,$c$ be positive reals such that $abc = 1$ . Prove that $\frac{a}{(a+3)^2} + \frac{b}{(b+3)^2} + \frac{c}{(c+3)^2} ≤ \frac{3}{16}$
What I've tried :
Well I was trying directly multiplying both sides by $(a+3)^2(b+3)^2(c+3)^2$ but it's too lengthy and seemed complex so any better way to solve it ?
Best Answer
This partial solution is too long to be a comment, and I think, it solves a large case of the problem. I hope someone can get it finished.
Lemma: If $x\leq y\leq z$, then we have: $$\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}\leq \frac{3}{2}.$$
Proof: WLOG, we assume $x=1$, hence $1\leq y \leq z$;
$$\frac{1}{1+y}+\frac{y}{y+z}+\frac{z}{z+1}\leq \frac{3}{2} \iff z^2(y-1)+z(1-y^2)+(y^2-y) \geq 0 \\ \iff (y-1)(z-y)(z-1)\geq 0,$$
which is obvious.
Now, let's assume $a=\frac{x}{y}, b=\frac{y}{z}, c=\frac{z}{x}$ such that $c \geq 1 \geq b \geq a$, which implies $x\leq y \leq z.$
We get:
$$\frac{a}{(a+3)^2}+\frac{b}{(b+3)^2}+\frac{c}{(c+3)^2}= \\ \frac{xy}{(x+3y)^2}+\frac{yz}{(y+3z)^2}+\frac{zx}{(z+3x)^2}= \\ \frac{xy}{x^2+6xy+9y^2}+\frac{yz}{y^2+6yz+9z^2}+\frac{zx}{z^2+6zx+9x^2} \leq \\ \frac{xy}{8xy+8y^2}+\frac{yz}{8yz+8z^2}+\frac{zx}{8zx+8x^2}=\frac{1}{8}(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}) \leq \frac{3}{16}.$$