# inequality with three variables and condition

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

• i think $a,b,c$ are positive – Dr. Sonnhard Graubner Jan 6 '16 at 18:08
• of course,i will change task now – chaos Jan 6 '16 at 18:09
• from where does this come? – Dr. Sonnhard Graubner Jan 6 '16 at 18:22
• If it is helpful, $\frac{3}{4}$ is attained where $a=b=c$. – CommonerG Jan 6 '16 at 20:57

## 1 Answer

By C-S and Vasc we obtain $\sum\limits_{cyc}\frac{a^2}{a^2+c}=\sum\limits_{cyc}\frac{a^4}{a^4+a^2c(a+b+c)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3c+a^2b^2+a^2bc)}\geq$

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

Id est, it remains to prove that $4(a^2+b^2+c^2)^2\geq\sum\limits_{cyc}(4a^4+5a^2b^2+3a^2bc)$, which is $\sum\limits_{cyc}c^2(a-b)^2\geq0$. Done!

• It is a good answer, but it is quite cryptical. I bet C-S is Cauchy-Schwarz, but what it Vasc? To improve the rendering would be a good idea, too, IMHO. – Jack D'Aurizio Jan 6 '16 at 20:01
• The Vasc;'s inequality it's the following. $(a^2+b^2+c^2)^2\geq3(a^3c+b^3a+c^3b)$. It's true because $(a^2+b^2+c^2)^2-3(a^3c+b^3a+c^3b)=\frac{1}{2}\sum\limits_{cyc}(a^2-b^2+ab-2ac+bc)^2\geq0$. – Michael Rozenberg Jan 6 '16 at 20:12