Prove the next cyclic inequality

Question

Let $a,b,c>0$ such that $$a+b+c=1$$

prove that

$$cyclic\sum \frac {ab}{\sqrt {c+ab}}\le\frac {1}{2} $$

By symetrie, i proved it by assuming $a=b=c$ but i cannot justify this hypothesis.

Answer 1

By C-S $$\left(\sum_{cyc}\frac{ab}{\sqrt{c+ab}}\right)^2=\left(\sum_{cyc}\frac{ab}{\sqrt{c(a+b+c)+ab}}\right)^2=\left(\sum_{cyc}\frac{ab}{\sqrt{(a+c)(b+c)}}\right)^2\leq$$ $$\leq\sum_{cyc}ab\sum_{cyc}\frac{ab}{(a+c)(b+c)}.$$ Thus, it's enough to prove that $$\sum_{cyc}ab\sum_{cyc}\frac{ab}{(a+c)(b+c)}\leq\frac{(a+b+c)^2}{4},$$ which is easy after full expanding.

Indeed, we need to prove that $$\sum_{cyc}(a^4b+a^4c-a^3b^2-a^3c^2-2a^3bc+2a^2b^2c)\geq0$$ or $$\sum_{cyc}(a-b)^2ab(a+b-c)\geq0.$$ Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}(a-b)^2ab(a+b-c)\geq$$ $$\geq(a-c)^2ac(a+c-b)+(b-c)^2bc(b+c-a)\geq$$ $$\geq(b-c)^2ac(a-b)+(b-c)^2bc(b-a)=(b-c)^2(a-b)^2c\geq0.$$ Done!