Proving that $f(t)=\frac{n^2}{2}\cdot t^{n-4}(1-t^2)\left(t^2-\frac{n-3}{n}\right)$ is bounded above by $1$, for $n\geq6$ and $t\in[0,1]$ 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).
 A: Ok, with some clever separation into cases I managed to do it.

*

*If $1-t^2 < \frac{2}{3n}$ then just bound:
$$f(t) < \frac{n^2}{2}\cdot 1^{n-4}\cdot \frac{2}{3n} \cdot \left( 1 - \frac{n-3}{n}\right) =1.$$


*If $1-t^2 \geq \frac{2}{3n}$, then $t^2\leq 1-\frac{2}{3n}$, so that:
$$f(t) \leq \frac{n^2}{2} t^{n-4} (1-t^2) \left(1-\frac{2}{3n} - \frac{n-3}{n} \right)= \frac{7n}{6} t^{n-4}(1-t)^2$$
and the maximum of $t^{n-4}(1-t^2)$ in $[0,1]$ can be found differentiating and equating to zero. It is exactly $\left(1-\frac{2}{n-2}\right)^{\frac{n-4}{2}} \frac{2}{n-2}$. So that, putting al this together:
$$f(t) \leq \frac{7}{3} \frac{n}{n-2} \left(1 - \frac{2}{n-2}\right)^{\frac{n-4}{2}}.$$
And as the term in the parentheses tends to $e^{-1} < 0.3679... < 3/7$, it is easy to conclude.
A: In case a less-clever estimate by separation of cases is useful:
Our goal is to show
$$
\frac{n^{2}}{2}\, t^{n-4}(1 - t^{2}) \left(t^{2} - \frac{n-3}{n}\right)
= \frac{n^{2}}{2}\, t^{n-4}(1 - t^{2}) \left(\frac{3}{n} - (1 - t^{2}\right)
\leq 1
$$
for $0 \leq t \leq 1$.
Write $n = 4 + m$ for $m \geq 2$ and $u = 1 - t^{2}$. The desired inequality becomes
$$
\tfrac{1}{2}(4 + m)^{2}\, (1 - u)^{m/2}\, u(\tfrac{3}{4 + m} - u) \leq 1
$$
for $0 \leq u \leq 1$. Put $c = \frac{3}{4 + m}$ and $v = \frac{u}{c}$, so the desired inequality reads
$$
\tfrac{9}{2}(1 - c v)^{m/2} (v - v^{2}) \leq 1
$$
for $0 \leq v \leq \frac{1}{c}$.
We have $v - v^{2} = \frac{2}{9} - (v - \frac{1}{3})(v - \frac{2}{3}) \leq \frac{2}{9}$ unless $\frac{1}{3} \leq v \leq \frac{2}{3}$. Since $0 \leq 1 - cv \leq 1$, the inequality in this case follows at once.
Otherwise (if $\frac{1}{3} \leq v \leq \frac{2}{3}$), we have $v - v^{2} = \frac{1}{4} - (v - \frac{1}{2})^{2} \leq \frac{1}{4}$ (for all real $v$), while
$$
(1 - cv)^{m/2} \leq (1 - \tfrac{1}{3}c)^{m/2} = (1 - \tfrac{1}{4 + m})^{m/2}.
$$
The right-hand side decreases with $m$, and for $m = 2$ is equal to $\frac{5}{6} < \frac{8}{9}$, so the inequality holds in this case as well.
