I have a problem that looks like a typical problem of maximizing functions in a compact interval. However, I am not being able to prove the bound I need.

Let $n\geq 6$ be an integer number. Consider the function: $$f(t) = \frac{n^2}{2} \cdot t^{n-4}(1-t^2) \left(t^2 - \frac{n-3}{n}\right) $$

Prove that for all $t\in [0,1]$ it holds $f(t) \leq 1$.

The points where the derivative $f'$ is zero are very ugly expressions.

By maximizing the factor $t^{n-4}(1-t^2)$ and using that the last factor is at most $\frac{3}{n}$ it is possible to deduce that for $n\geq 6$ it is $f(t) \leq \frac{3}{2}$ (in fact, it is possible to bound it in the limit by $\frac{3}{e}\approx 1.1036...$ but that is far from $1$.

I have checked that the claim is true for several random values of $n$ (in fact, I think that the bound can be reduced to something like $0.61...$ for $n>20$ say).