<s>[INST] Hard Olympiad Inequality 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.
 [/INST] 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!
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