Showing that $g(t)=t\left(1-\frac{t}{2}\right)^n \leq \frac{2}{n+1}$ for every natural $n$ and $t$ in $[0, 1]$.
How is this done? Is there a simple way to prove this? I tried putting in numbers and it seems to work out every time. I tried multiplying both sides by (n+1). I don't know what to do next.
As Kelenner suggested, compute the derivative $$g(t)=t\Big(1-\frac{t}{2}\Big)^n $$ $$g'(t)=\Big(1-\frac{t}{2}\Big)^{n-1}\Big(1-\frac{n+1}{2}t\Big)$$ So, $g'(t)=0$ when $t=\frac{2}{n+1}$ and then $$g\Big(\frac{2}{n+1}\Big)=\frac{2 \left(1-\frac{1}{n+1}\right)^n}{n+1}$$
I am sure that you can take from here.
An approach without calculus.
Since for $t=0$ the inequality is trivially satisfied, rewrite it as $$\left(\frac{2-t}{2}\right)^n\le\frac{2}{t(n+1)} $$ $$\ \ (2-t)^n\le\frac{2^{n+1}}{t(n+1)}.\tag{P(n)}$$ We first prove by induction that P$(n)$ holds for $\displaystyle t\ge \frac{2}{n+2}$, and then we show otherwise it holds also for $\displaystyle t< \frac{2}{n+2}.$
Basis: P$(0)$: $\displaystyle 1\le \frac{2}{t}$ is true, given that $t\le1$.
Inductive step: Assume P$(k)$ holds. P$(k+1)$ is equivalent to$$(2-t)^k(2-t)\le \frac{2^{k+1}}{t(k+1)}\frac{2(k+1)}{k+2}, $$ so it holds too if $$2-t\le \frac{2(k+1)}{k+2} \\ 2-t \le 2-\frac{2}{k+2} \\ t\ge \frac{2}{k+2}>\frac{2}{(k+1)+2}.$$
As $t>0,$ we have $$(2-t)^n<2^n,$$ thus write $$2^n\le \frac{2^{n+1}}{t(n+1)}$$$$1\le \frac{2}{t(n+1)}.$$ As long as we're considering $\displaystyle t<\frac{2}{n+2}$ indeed we have $$1<\frac{n+2}{n+1}<\frac{2}{t(n+1)},$$ and thus P$(n)$ is proven for all $n$ and all $t$ in $[0, 1]$.
