Let $a,b,c>0$ and $a+b+c=3$. Find the minimum of $P$. Let $a,b,c>0$ and $a+b+c=3$. Prove that: 
$$P=\frac{{a}^{2}}{\sqrt{{b}^{3}+1}}+ \frac { { b }^{ 2 } }{ \sqrt{ { c }^{ 3 }+1 } }+\frac{{c}^{2}}{\sqrt{{a}^{3}+1}}\geq \frac{3\sqrt{2}}{2}$$
 A: By AM-GM, C-S and Rearrangement we obtain:
$$\sum_{cyc}\frac{a^2}{\sqrt{b^3+1}}=\sum_{cyc}\frac{2\sqrt2a^2}{2\sqrt{2(b^2-b+1)(b+1)}}\geq\sum_{cyc}\frac{2\sqrt2a^2}{2(b^2-b+1)+(b+1)}=$$
$$=\sum_{cyc}\frac{2\sqrt2a^2}{2b^2-b+3}=\sum_{cyc}\frac{2\sqrt2a^4}{2a^2b^2-a^2b+3a^2}\geq\frac{2\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2-a^2b+3a^2)}=$$
$$=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-3a^2b+9a^2)}=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-(a+b+c)a^2b+(a+b+c)^2a^2)}=$$
$$=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-a^3b-a^2b^2-a^2bc+a^4+2a^2b^2+2a^3b+2a^3c+2a^2bc)}=$$
$$=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3b+2a^3c+7a^2b^2+a^2bc)}\geq\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)}.$$
Thus, it remains to prove that
$$\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)}\geq\frac{3\sqrt2}{2}$$
or
$$4(a^2+b^2+c^2)^2\geq\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)$$ or
$$\sum_{cyc}(2a^4-a^3b-a^3c+a^2b^2-a^2bc)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a^2+ab+b^2)+\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
Done!
