Want help with proving a calculus theorem

Let $f(x)$ have a second derivative on the closed interval $[-2,2]$. If $left| f(x) right| le 1$ and $frac{1}{2} (f^{prime}(0))^2+f(0)^3>frac{3}{2} $ when $-2le xle2$, now I need to prove that there must be a point $x_{0}$ on the interval $(-2,2)$ such that $f^{prime prime}left(x_{0}right)+3fleft(x_{0}right)^2=0$.

(Series[1/2 (f'[x])^2 + f[x]^3, {x, 0, 1}]) // FullSimplify

1/2 (Series[f'[x], {x, 0, 1}] // Normal)^2 + (Series[
     f[x], {x, 0, 1}] // Normal)^3 // Expand

(Series[f''[x], {x, 0, 1}] // Normal)^2 + (3*Series[f[x], {x, 0, 1}] //
     Normal)^2 // Expand

The above code does not reveal the nature of the problem and solve it cleverly. What can I do to solve this problem?

In addition, I really want to use MMA to solve this problem and deepen my understanding of such problems with MMA. Please do not close this post for the time being.

The source of this problem (张宇高等数学18讲):