Prove that $0.5x^2 -3x ≥ -4.5$ for all real numbers x.

Question

I'm not familiar at all with inequality proofs. How do I approach this problem?

Answer 1

Let $f(x)=0.5x^2-3x+4.5$

$$f(x)=0.5(x^2-6x+9)=0.5(x-3)^2\ge0$$ for all real $x$.

Thus, $$0.5x^2-3x\ge-4.5$$

Answer 2

$$0.5x^2-3x\ge -4.5\iff x^2-6x\ge -9\iff x^2-6x+9\ge 0=(x-3)^2\ge 0$$ This last inequality is evident.

Answer 3

Just another way: by AM-GM inequality $0.5x^2+4.5\geqslant 2\sqrt{2.25x^2}=3|x|\geqslant 3x$.