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: $$\frac{3}{4}\left(\frac{a}{b}+\frac{b}{a}\right)-\frac{1}{2}-\frac{3a^2-2ab+3b^2}{(a+b)^2}=$$ $$=\frac{3a^2-2ab+3b^2}{4ab}-\frac{3a^2-2ab+3b^2}{(a+b)^2}=\frac{(a-b)^2(3a^2-2ab+3b^2)}{4ab(a+b)^2}\geq0.$$

  • $\begingroup$ 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 $\endgroup$
    – Adele
    May 12 '20 at 11:54
  • $\begingroup$ Can you help me please? $\endgroup$
    – Adele
    May 12 '20 at 11:55
  • $\begingroup$ @Adele I added something. See now. $\endgroup$ May 12 '20 at 12:01
  • $\begingroup$ I have seen, thank you very much! $\endgroup$
    – Adele
    May 12 '20 at 13:20
  • $\begingroup$ @Adele You are welcome! $\endgroup$ May 12 '20 at 13:20

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