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.                         
 A: Let $a+2b+c = x $ etc.
$$
f(a,b,c)=\frac{x+y+z}{4}-\frac{9(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})+\frac{9z}{(x)^2}+\frac{9x}{(y)^2}+\frac{9y}{(z)^2}
$$
By the rearrangement inequality, 
$\frac{9z}{(x)^2}+\frac{9x}{(y)^2}+\frac{9y}{(z)^2}\ge \frac{9x}{(x)^2}+\frac{9y}{(y)^2}+\frac{9z}{(z)^2} $ as the minimum occurs when numerators and denominators are sorted in the same order. So we have 
$$
f(x,y,z)\ge\frac{x+y+z}{4}+ 9(\frac1{x}+\frac1{y}+\frac1{z})-\frac{9(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})
$$
Now this is a symmetric convex function in the variables, where the minimum occurs when all variables equal, i.e.
$$
f(x,y,z)\ge\frac{3x}{4}+ \frac{27}{x}-\frac{81}{4 x} = \frac{3x}{4}+ \frac{27}{ 4 x} = \frac{9}{4}(\frac{x}3 + \frac3{x})
$$
The minimum of this occurs at $x=3$ so
$$
f(x,y,z)\ge \frac{9}{2}
$$
Now indeed that happens, as you already computed, at $a=b=c=3/4$ (which corresponds to $x=y=z=3$), so the bound is tight and this is your minimum.

Considering your last inequality, you can proceed in the same fashion:
$$
\frac{a}{(a+2b+c)^2}+\frac{b}{(b+2c+a)^2}+\frac{c}{(c+2a+b)^2} \geq \frac{9}{16(a+b+c)}\\
\leftrightarrow 
-\frac{(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})+\frac{z}{(x)^2}+\frac{x}{(y)^2}+\frac{y}{(z)^2}- \frac{9}{4(x+y+z)}  \geq 0
$$
Now the LHS can be bound by rearrangement, and you have  the stronger condition
$$ 
-\frac{(x+y+z)}{4}(\frac{1}{(x)^2}+\frac{1}{(y)^2}+\frac{1}{(z)^2})+\frac1{x}+\frac1{y}+\frac1{z}- \frac{9}{4(x+y+z)}  \geq 0
$$
Again using symmetry, an even stronger condition is
$$ 
-\frac{9}{4 x}+\frac3{x}- \frac{3}{4x}  \geq 0
$$
but the LHS is identically zero and so this holds. Again, verify that  $a=b=c=3/4$ solves the starting inequality, which says that the starting inequality is tight.
A: 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.
