Hard Olympiad Inequality

Question

Let x,y,z be positive real numbers such that $xy+xz+yz=1$. Prove that $$\sqrt{x^3+x}+ \sqrt{y^3+y}+ \sqrt{z^3+z} \geq 2 \cdot \sqrt{x+y+z}$$. I tried to square expand homogenize then majorize. But I couldn't make it work. Any help would be much appreciated.

Answer 1

problem: since $a,b,c>0$,and such $ab+bc+ac=1$, show that $$\sqrt{a^3+a}+\sqrt{b^3+b}+\sqrt{c^3+c}\ge 2\sqrt{a+b+c}$$

Poof: Using Holder inequality,we have $$\left(\sum\sqrt{a^3+a}\right)^2\left(\sum\dfrac{a^2}{a^2+1}\right)\ge\left(\sum a\right)^3$$ it remains to prove that $$\left(\sum a\right)^2\ge 4\sum\dfrac{a^2}{a^2+1}$$

which is true,because \begin{align*}\left(\sum a\right)^2- 4\sum\dfrac{a^2}{a^2+1}&=\left(\sum a\right)^2-4\left(\sum bc\right)\left[\sum\dfrac{a^2}{(c+a)(a+b)}\right]\\ &=\sum\dfrac{a(b-c)^2(b+c-a)^2}{(b+a)(b+c)(a+c)}\\ &\ge 0 \end{align*}

Answer 2

By C-S and Schur we obtain: $$\sum_{cyc}\sqrt{a^3+a}=\sqrt{\left(\sum_{cyc}\sqrt{a^3+a^2b+a^2c+abc}\right)^2}=$$ $$=\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}(a^3+a^2b+a^2c+abc+2(ab(a+b+c)+abc))}=$$ $$=\sqrt{\sum_{cyc}(a^3+3a^2b+3a^2c+5abc)}=\sqrt{\sum_{cyc}(a^3-a^2b-a^2c+abc+4(a^2b+a^2c+abc))}\geq$$ $$\geq\sqrt{4\sum_{cyc}(a^2b+a^2c+abc)}=\sqrt{4(a+b+c)(ab+ac+bc)}=2\sqrt{a+b+c}.$$ Done!