 kixEffinsoj

2022-06-22

inequality with three variables and condition
If $a$, $b$ and $c$ positive real numbers such that $a+b+c=1$, prove
$\frac{{b}^{2}}{a+{b}^{2}}+\frac{{c}^{2}}{b+{c}^{2}}+\frac{{a}^{2}}{c+{a}^{2}}⩾\frac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea? laure6237ma

By C-S and Vasc we obtain $\sum _{cyc}\frac{{a}^{2}}{{a}^{2}+c}=\sum _{cyc}\frac{{a}^{4}}{{a}^{4}+{a}^{2}c\left(a+b+c\right)}\ge \frac{\left({a}^{2}+{b}^{2}+{c}^{2}{\right)}^{2}}{\sum _{cyc}\left({a}^{4}+{a}^{3}c+{a}^{2}{b}^{2}+{a}^{2}bc\right)}\ge$
$\ge \frac{\left({a}^{2}+{b}^{2}+{c}^{2}{\right)}^{2}}{\sum _{cyc}\left({a}^{4}+{a}^{2}{b}^{2}+{a}^{2}bc\right)+\frac{\left({a}^{2}+{b}^{2}+{c}^{2}{\right)}^{2}}{3}}=\frac{3\left({a}^{2}+{b}^{2}+{c}^{2}{\right)}^{2}}{\sum _{cyc}\left(4{a}^{4}+5{a}^{2}{b}^{2}+3{a}^{2}bc\right)}$
Id est, it remains to prove that $4\left({a}^{2}+{b}^{2}+{c}^{2}{\right)}^{2}\ge \sum _{cyc}\left(4{a}^{4}+5{a}^{2}{b}^{2}+3{a}^{2}bc\right)$, which is
$\sum _{cyc}{c}^{2}\left(a-b{\right)}^{2}\ge 0$
Done!

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