Let $a,b,c$ be positive reals. Find the minimum value of $P=\frac{a^2}{(a+b)^2}+\frac{b^2}{(b+c)^2}+\frac{c}{4a}$. This inequality problem is from Phan Boi Chau High School for the Gifted Grade 10 Selection Test, Nghe An, Vietnam.
Let $a,b,c$ be positive reals. Find the minimum value of $P=\dfrac{a^2}{(a+b)^2}+\dfrac{b^2}{(b+c)^2}+\dfrac{c}{4a}$.
I've done $\dfrac{c}{4a}=\dfrac{c^2}{4ac}\geq\dfrac{c^2}{(c+a)^2}$.
I think the minimum value is $\dfrac34$, so the problem is to prove $\displaystyle \sum_\text{cyc}\dfrac{a^2}{(a+b)^2}\geq\dfrac34$. But I still can't.
Can you help?
 A: Yes, you are right!
The minimal value is $\frac{3}{4}$ and occurs for $a=b=c$.
We can rewrite this inequality in the following form:
$$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}+\frac{1}{4z}\geq\frac{3}{4},$$
where $\frac{b}{a}=x$, $\frac{c}{b}=y$ and $\frac{a}{c}=z$.
Thus, $xyz=1$ and since $$\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}\geq\frac{1}{1+xy}$$ it's $$xy(x-y)^2+(xy-1)^2\geq0,$$ it's enough to prove that
$$\frac{1}{1+\frac{1}{z}}+\frac{1}{4z}\geq\frac{3}{4},$$ which is
$$(z-1)^2\geq0.$$
Done! 
It's interesting that your trying  also helps. 
Indeed, it's enough to prove that
$$\sum_{cyc}\frac{a^2}{(a+b)^2}\geq\frac{3}{4}.$$
Now, by C-S we obtain:
$$\sum_{cyc}\frac{a^2}{(a+b)^2}=\sum_{cyc}\frac{a^2(a+c)^2}{(a+b)^2(a+c)^2}\geq\frac{\left(\sum\limits_{cyc}(a^2+ab)\right)^2}{\sum\limits_{cyc}(a+b)^2(a+c)^2}.$$
Thus, it remains to prove that
$$4\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a+b)^2(a+c)^2$$ or
$$\sum_{cyc}(a^4+2a^3b+2a^3c+3a^2b^2-8a^2bc)\geq0,$$
which is true by Muirhead because $(4,0,0)\succ(2,1,1)$, $(3,1,0)\succ(2,1,1)$ and $(2,2,0)\succ(2,1,1)$.
