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

The only thing I have tried till now is graphing, from which I proved this, but how would you prove this mathematically?
\begin{align*}
\left(x+\frac{1}{2}\right)^3-3x^2-\frac{1}{8}&amp;=x^3-\frac{3x^2}{2}+\frac{3x}{4}\\
&amp;=x\left(x-\frac{3}{4}\right)^2+\frac{3x}{16}\\
&amp;&gt;0
\end{align*}
$$(x+0.5)^3-3x^2-0.125=x^3+1.5x^2+0.75x+0.125=x^3-1.5x^2+0.75x$$

which is then equal to

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

which is positive for all $0&lt;x$.