Prove that when $x>0$, $(x+1/2)^3>3x^2+1/8$ without graphing

Question

Prove that when $x>0$, $(x+1/2)^3>3x^2+1/8$ without graphing.

The only thing I have tried till now is graphing, from which I proved this, but how would you prove this mathematically?

Answer 1

\begin{align*} \left(x+\frac{1}{2}\right)^3-3x^2-\frac{1}{8}&=x^3-\frac{3x^2}{2}+\frac{3x}{4}\\ &=x\left(x-\frac{3}{4}\right)^2+\frac{3x}{16}\\ &>0 \end{align*}

Answer 2

Let $f(x)=\bigg(x+\cfrac 12\bigg)^3-3x^2-\cfrac 18$.

$$f(x)=x^3-\cfrac 32 x^2+\cfrac 34x $$

$$f'(x)=3x^2-3x+\cfrac 34$$

$$f'(x)=3\bigg(x-\cfrac 12\bigg)^2$$

What can you tell about $x$ from here?

It is always monotone increasing.

From $f(0)=0, x>0$ and $f(x)>0$. Therefore, when $x>0$, $\left(x+\cfrac 12\right)^3>3x^2+\cfrac 18$

Answer 3

$$(x+0.5)^3-3x^2-0.125=x^3+1.5x^2+0.75x+0.125=x^3-1.5x^2+0.75x$$

which is then equal to

$$x(x^2-1.5x+0.75)=x((x-0.75)^2+0.9375)$$

which is positive for all $0<x$.