TITLE: Prove inequality with powers of $n$
QUESTION [5 upvotes]: Prove that $(a+b)^n \le a^n+b((a+1)^n-a^n)$,
n in N,
$a \ge 0, 0<b<1$,
$a,b$ in $R$.,
With the little I remember from high school and college, I have tried the binomial expansion of $(a+1)^n$ and $(a+b)^n$.
Also tried to move $a^n$ at the LHS and use the formula:
$x^n-y^n=(x-y)\left(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\dots+x^2y^{n-3}+xy^{n-2}+y^{n-1}\right)$ but I am not getting anywhere.
FYI this is not homework or anything. Just trying to keep my brain alive, to minimize the risk of developing dementia :)
Thank you very much in advance!!

REPLY [4 votes]: Let $f(x) = x^n$. Since this is convex, so
$$\dfrac{f(x+b)-f(x)}{b}\le \dfrac{f(x+1)-f(x)}{1}\qquad \forall b\in(0,1)$$
Hence the result.
Consider the following image:

For convex red curve, the slope of blue secant is more than that of the green one. In general, slope of secant joining points $x$ and $x+y$ increases with $y$, where $y>0$.