<s>[INST] Inequality for optimization. I found this problem:
Find the minimum value of:
$f(a,b,c)=a+b+c+\frac{9a}{(a+2b+c)^2}+\frac{9b}{(b+2c+a)^2}+\frac{9c}{(c+2a+b)^2}$
Where $a,b,c >0$.
After looking for some minimum values, I think the lowest one is for the equality case $a=b=c=3/4$.Obviously,
$f(a,a,a)=3a+\frac{27}{16a} \geq 2\sqrt{\frac{81}{16}}=9/2$
I got inspired from the equality case to "substitute" $a=(a+b+c)/3$ and then having this inequality:
$f(a,b,c) \geq 3(\frac{a+b+c}{3})+\frac{27}{16((a+b+c)/3)}=3(\frac{a+b+c}{3})+\frac{81}{16(a+b+c)}$.
Subtracting on the both sides by $a+b+c$ and simplifying by $9$, we get
$\frac{a}{(a+2b+c)^2}+\frac{b}{(b+2c+a)^2}+\frac{c}{(c+2a+b)^2} \geq \frac{9}{16(a+b+c)}$
This last inequality is really the problem
 I`m pretty sure it is true after giving some arbitrary values and also looking at cases when some variables tend to infinity.
Still, I cannot prove it. I tried Titu's lemma but end up with an invalid inequality.
I'd like to see a nice proof for this inequality.                         
 [/INST] By AM-GM and C-S we obtain:
$$\sum_{cyc}\left(a+\frac{9}{(a+2b+c)^2}\right)\geq6\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{(a+2b+c)^2}}\geq6\sum_{cyc}\frac{a}{a+2b+c}=$$
$$=6\sum_{cyc}\frac{a^2}{a(a+2b+c)}\geq\frac{6(a+b+c)^2}{\sum\limits_{cyc}a(a+2b+c)}=\frac{6(a+b+c)^2}{\sum\limits_{cyc}(a^2+3ab)}\geq\frac{9}{2},$$ where the last ineqyality it's just $\sum\limits_{cyc}(a-b)^2\geq0.$
The equality occurs for $a=b=c=\frac{3}{4},$ which says that we got a minimal value.
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