If $a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$, then $a^2+b^2 \ge 2$ Prove that if
$$a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$$
then $$a^2+b^2 \ge 2.$$

From the conditions we have $\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}\geq2$.
Thus, it remains to prove that $a^2+b^2\geq\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}$, which is $a^2b^2\geq0$.
Done!
