Prove that: $a+b+c+a^2+b^2+c^2\ge\sum_{cyc}{\sqrt{a(a^3+b+c)}}$

Question

Given $a,b,c$ be non-negative real numbers such that: $ab+bc+ca+2abc=1.$ Prove that: $$a+b+c+a^2+b^2+c^2\ge\sqrt{a(a^3+b+c)}+\sqrt{b(b^3+c+a)}+\sqrt{c(c^3+a+b)}$$ I tried AM-GM for right side: $$\sum_{cyc}{\sqrt{a(a^3+b+c)}}\le\frac{1}{3}\sum_{cyc}{\left(\frac{9}{4}a+b+c+a^3\right)}=\frac{a^3+b^3+c^3}{3}+\frac{17(a+b+c)}{12}$$ But it seems i am lost. Also, the condition is not relevant!. I hope we can find a good way. Thank for sharing your knowledge and your time.

Answer 1

Let $a=\frac{x}{y+z}$ and $b=\frac{y}{x+z},$ where $x$, $y$ and $z$ are positives.

Thus, by the condition $c=\frac{z}{x+y}.$

Now, by C-S $$\sum_{cyc}\sqrt{a(a^3+b+c)}\leq\sqrt{\sum_{cyc}a\sum_{cyc}(a^3+b+c)}.$$ Thus, it's enough to prove that: $$\left(\sum_{cyc}(a^2+a)\right)^2\geq\sum_{cyc}a\sum_{cyc}(a^3+2a)$$ or $$(a+b+c)\sum_{cyc}(2a^2-a)\geq\sum_{cyc}(a^3b+a^3c-2a^2b^2)$$ or $$(a+b+c)\sum_{cyc}\left(\frac{2x^2}{(y+z)^2}-\frac{x}{y+z}\right)\geq\sum_{cyc}ab(a-b)^2$$ or $$(a+b+c)\sum_{cyc}\frac{x((x-y-(z-x))}{(y+z)^2}\geq\sum_{cyc}\frac{xy}{(y+z)(x+z)}\left(\frac{x}{y+z}-\frac{y}{x+z}\right)^2$$ or $$(a+b+c)\sum_{cyc}(x-y)\left(\frac{x}{(y+z)^2}-\frac{y}{(x+z)^2}\right)\geq\sum_{cyc}\frac{xy(x-y)^2(x+y+z)^2}{(y+z)^3(x+z)^3}$$ or $$\sum_{cyc}(x-y)^2\left(\frac{(a+b+c)(x^2+y^2+z^2+xy+2xz+2yz)}{(y+z)^2(x+z)^2}-\frac{xy(x+y+z)^2}{(y+z)^3(x+z)^3}\right)\geq0$$ and since by Nesbitt $$a+b+c\geq\sum_{cyc}\frac{x}{y+z}\geq\frac{3}{2},$$ it's enough to prove that: $$3(y+z)(x+z)(x^2+y^2+z^2+xy+2xz+2yz)\geq2xy(x+y+z)^2,$$ which is true because $$3(y+z)(x+z)(x^2+y^2+z^2+xy+2xz+2yz)\geq$$ $$\geq3xy(x^2+y^2+z^2+xy+2xz+2yz)\geq2xy(x+y+z)^2.$$