# How do I show that $\sum_{\text{cyc}}\frac{3a^2-2ab+3b^2}{(a+b)^2} < \frac{9}{4}(\frac{a}{c} + \frac{c}{a}) - \frac{3}{2}$?

Let the real numbers $$a,b,c \in \mathbb(0, \infty\ )$$, $$a\leq b\leq c$$. Prove that $$\frac{3a^2-2ab+3b^2}{(a+b)^2}+\frac{3b^2-2bc+3c^2}{(b+c)^2}+ \frac{3c^2-2ac+3a^2}{(c+a)^2} \leq\frac{9}{4}(\frac{a}{c}+\frac{c}{a}) -\frac{3}{2}.$$ I tried to prove that $$\frac{3a^2-2ab+3b^2}{(a+b)^2}\leq\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2},$$ but it didn't work.

By your work $$\sum_{cyc}\frac{3a^2-2ab+3b^2}{(a+b)^2}\leq\sum_{cyc}\left(\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2}\right)\leq\frac{9}{4}\left(\frac{a}{c}+\frac{c}{a}\right)-\frac{3}{2}$$ because $$\frac{a}{b}+\frac{b}{a}\leq\frac{a}{c}+\frac{c}{a}$$ (it's just $$bc-a^2\geq0$$) and $$\frac{b}{c}+\frac{c}{b}\leq\frac{a}{c}+\frac{c}{a}$$ (it's just $$c^2-ab\geq0$$).
The first inequality we can prove by the following way: $$\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2}-\frac{3a^2-2ab+3b^2}{(a+b)^2}=$$ $$=\frac{3a^2-2ab+3b^2}{4ab}-\frac{3a^2-2ab+3b^2}{(a+b)^2}=\frac{(a-b)^2(3a^2-2ab+3b^2)}{4ab(a+b)^2}\geq0.$$
• Thank you but I haven't proved that $$\frac{3a^2-2ab+3b^2}{(a+b)^2}\leq\frac{3}{4}(\frac{a}{b}+\frac{b}{a}) -\frac{1}{2}$$.I said that I tried to prove this but I failed May 12 '20 at 11:54