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

\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*}
