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

$\frac{{b}^{2}}{a+{b}^{2}}}+{\displaystyle \frac{{c}^{2}}{b+{c}^{2}}}+{\displaystyle \frac{{a}^{2}}{c+{a}^{2}}}\u2a7e{\displaystyle \frac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?

If $a$, $b$ and $c$ positive real numbers such that $a+b+c=1$, prove

$\frac{{b}^{2}}{a+{b}^{2}}}+{\displaystyle \frac{{c}^{2}}{b+{c}^{2}}}+{\displaystyle \frac{{a}^{2}}{c+{a}^{2}}}\u2a7e{\displaystyle \frac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?

laure6237ma

Beginner2022-06-23Added 27 answers

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(a+b+c)}\ge \frac{({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}}{\sum _{cyc}({a}^{4}+{a}^{3}c+{a}^{2}{b}^{2}+{a}^{2}bc)}\ge $

$\ge \frac{({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}}{\sum _{cyc}({a}^{4}+{a}^{2}{b}^{2}+{a}^{2}bc)+\frac{({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}}{3}}=\frac{3({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}}{\sum _{cyc}(4{a}^{4}+5{a}^{2}{b}^{2}+3{a}^{2}bc)}$

Id est, it remains to prove that $4({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}\ge \sum _{cyc}(4{a}^{4}+5{a}^{2}{b}^{2}+3{a}^{2}bc)$, which is

$\sum _{cyc}{c}^{2}(a-b{)}^{2}\ge 0$

Done!

$\ge \frac{({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}}{\sum _{cyc}({a}^{4}+{a}^{2}{b}^{2}+{a}^{2}bc)+\frac{({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}}{3}}=\frac{3({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}}{\sum _{cyc}(4{a}^{4}+5{a}^{2}{b}^{2}+3{a}^{2}bc)}$

Id est, it remains to prove that $4({a}^{2}+{b}^{2}+{c}^{2}{)}^{2}\ge \sum _{cyc}(4{a}^{4}+5{a}^{2}{b}^{2}+3{a}^{2}bc)$, which is

$\sum _{cyc}{c}^{2}(a-b{)}^{2}\ge 0$

Done!