$\frac{a(a+b)}{4a^2+ab+b^2} + \frac{b(b+c)}{4b^2+bc+c^2} + \frac{c(c+a)}{4c^2+ca+a^2} \leq 1$

Question

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!

Answer 1

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.