rightways wrote: For positive reals $a,b,c$ with $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 3$ prove that $$\frac{a^3}{a^2+b}+\frac{b^3}{b^2+c}+\frac{c^3}{c^2+a}\ge \frac{3}{2}$$ $$\sum a\geq\frac{1}{3}(\sum \sqrt{a})^2\geq \sum \sqrt{a}$$$$\sum \frac{a^3}{a^2+b}=\sum a-\sum \frac{ab}{a^2+b}\geq \sum a-\frac{1}{2}\sum \sqrt{a}\geq \frac{1}{2}\sum \sqrt{a}\geq \dfrac {3} {2}.$$ https://artofproblemsolving.com/community/c6h545390p3158196

For positive reals $a,b,c$ with $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 3$ prove that $$\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge \frac{3}{2}$$$$\frac{a^3}{ka^2+b}+\frac{b^3}{kb^2+c}+\frac{c^3}{kc^2+a}\geq \dfrac {3} {k+1}$$Where $k\in N^+.$

rightways wrote: For positive reals $a,b,c$ with $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 3$ prove that $$\frac{a^3}{a^2+b}+\frac{b^3}{b^2+c}+\frac{c^3}{c^2+a}\ge \frac{3}{2}$$

Attachments:

For positive reals $a,b,c$ with $\sqrt{a}+\sqrt{b}+\sqrt{c}\ge 3$ prove that $$\frac{a^k}{a^2+b}+\frac{b^k}{b^2+c}+\frac{c^k}{c^2+a} \ge \frac{3}{2}$$Where $3\leq k\in N^+.$