We are given that $f(a)-a\ge f(b)-b$ for every $a,b>0$. Furthermore, taking $a:=x^2$ we also have $f(a)-a^3 \ge f(b)-b^3$. Under these, our goal is to show \[ f(a) - f(b) \ge \frac{\sqrt{3}}{2}(a-b)(a+b). \]Suppose first $a+b\le 2/\sqrt{3}$. Then, $f(a)-f(b)\ge a-b\ge (\sqrt{3}/2)(a-b)(a+b)$, completing the proof. Suppose next $a+b>2/\sqrt{3}$. We have that $4(a^2+ab+b^2)\ge 3(a+b)^2$, so $f(a)-f(b)\ge (a-b)(a^2+ab+b^2) \ge (3/4)(a-b)(a+b)^2 >(\sqrt{3}/2)(a-b)(a+b)$, completing the proof.