Multiply the condition with $b^2c^2$ to get $bc(1-c^2) \geq 0$. Then the inequality follows by summing: $$a^2+b^2\geq 2|ab|\geq |ab| \geq ab$$ $$5\left(\frac{b^2}{4}+c^2-bc^3\right) \geq 5(|bc|-bc^3) \geq 5(bc-bc^3)=5bc(1-c^2)\geq 0$$ $$4a^2+\frac{11b^2}{4}\geq 0$$

For real numbers, $a,b,c$ with $bc \ne 0$ we have to $\frac{1-c^2}{bc} \ge 0$. Prove that $$a^2+b^2+c^2 -bc^3 \ge \sqrt 3 ab $$ parmenides51 wrote: For real numbers, $a,b,c$ with $bc \ne 0$ we have to $\frac{1-c^2}{bc} \ge 0$. Prove that $$5( a^2+b^2+c^2 -bc^3) \ge ab.$$