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 $ -2\le x\le2$ , 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)+3\left[f\left(x_{0}\right)\right]=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?