<s>[INST] $\frac{a(a+b)}{4a^2+ab+b^2} + \frac{b(b+c)}{4b^2+bc+c^2} + \frac{c(c+a)}{4c^2+ca+a^2} \leq 1$ I've got stuck at this problem:

Let $a$, $b$, $c$ be real numbers. Prove that
  $$\frac{a(a+b)}{4a^2+ab+b^2} + \frac{b(b+c)}{4b^2+bc+c^2} + \frac{c(c+a)}{4c^2+ca+a^2} \leq 1$$

Firstly, I've thought this :
$$a^2 - 2ab + b^2 \geq 0$$
$$4a^2 + ab + b^2 \geq 3(ab + a^2)$$
$$\frac{a(a+b)}{4a^2 + ab + b^2} \leq \frac{1}{3}$$
Similarly we obtain that
$$\frac{b(b+c)}{4b^2+bc+c^2} \leq \frac{1}{3}$$
And 
$$\frac{c(c+a)}{4c^2+ca+a^2} \leq \frac{1}{3}$$
Summing all, we obtain the inequality.
Is this way correct?
(I have doubts about my solution because this problem was found in a math magazine which usually has difficult problems(at least for me)). Or is there a another way?
Thanks!
 [/INST] The inequality is equivalent to:
$$
\sum_{cyc}\frac{3a(a+b)}{4a^2+ab+b^2}≤3\iff \\
0≤\sum_{cyc}1-\frac{3a(a+b)}{4a^2+ab+b^2}=\sum_{cyc}\frac{a^2-2ab+b^2}{4a^2+ab+b^2}=\sum_{cyc}\frac{(a-b)^2}{(a-b)^2+3a(a+b)}\iff\\
0≤\sum_{cyc}\frac{(a-b)^2}{\left(\frac{7}{4}a-\frac{1}{4}b\right)^2+\frac{15}{16}(a+b)^2}
$$
Which is obvious.
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