If $A$ and $B$ are $n\times n$ matrices, prove the inequality $$\Vert(A+B)^n-A^n\Vert\le\Vert B\Vert\Vert A\Vert^{n-1}e^{\frac{\Vert B\Vert}{\Vert A\Vert}}.$$
I would be very happy if you could show me a way to solve this problem.
If $A$ and $B$ are $n\times n$ matrices, prove the inequality $$\Vert(A+B)^n-A^n\Vert\le\Vert B\Vert\Vert A\Vert^{n-1}e^{\frac{\Vert B\Vert}{\Vert A\Vert}}.$$
I would be very happy if you could show me a way to solve this problem.
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The inequality is not true in general. Take $$ A_0=\begin{bmatrix}0&1\\0&0\end{bmatrix},\ \ \ B_0=\begin{bmatrix}2&0\\0&0\end{bmatrix}. $$ Now form $A,B$, of size $n\times n$, for $n>2$, by putting $$A=\begin{bmatrix}A_0&0\\0&0\end{bmatrix}, \ \ \ B=\begin{bmatrix}B_0&0\\0&0\end{bmatrix}.$$ Then $$ A^n=0,\ \ (A+B)^n-A^n=(A+B)^n=2^n(A+B). $$ So, in any norm where $\|A\|=1$ (all of the common ones), the left hand side of your inequality goes to infinity with $n$, while the right-hand-side is fixed. So, for any of those norms, there exists $n$ big enough such that the inequality fails.
When $AB=BA$ and a matricial norm (i.e., submultiplicative) though, the situation is different, because the binomial formula applies. I don't see an obvious way to prove the inequality, though.