I want to prove that that $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ is true with the help of induction.
base case: for $n=0$ we get $(a+b)^0 \leq 2^{-1}(a^0 +b^0) \Longleftrightarrow 1 \leq 1$. Thus the inequality for $n=0$ is correct.
IH: For a any $n \in \mathbb{N}$ and $a,b \geq 0$, $(a+b)^n \leq 2^{n-1...
From a logic standpoint, you should not be working both sides of the inequality at once. You should be starting with, say $(a+b)^{n+1}$ and then show that it is less or equal to $2^n(a^{n+1}+b^{n+1})$.
instead of $=$ you should use $\Longleftrightarrow$ to indicate, that the equation is equivalent to the next equation. Otherwise it is misleading, because one could think that the expression in the next line after the = was equal to the expression on the right-hand side equation from before
From How to ask a good question Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title.
Avoid the use of * to denote multiplication. That's usual in programming, not in Mathematics.
In a step apparently you used $(a^n+b^n)(a+b) = a^n a+b^n b$ which is false.
Actually, what you've just written is correct, but you would get $2^n(a^{n+1}+b^{n+1}+a^nb+ab^n)\leq 2^n(a^{n+1}+b^{n+1}+a^{n+1}+b^{n+1})=2^{n+1}(a^{n+1}+b^{n+1})$, so you would have to divide both sides by $2$ again to get your desired result. Basically, you should not have multiplied by $2$ at all.
Yes, you do. Otherwise the proof is unfinished. You need to show that $(a+b)^{n+1}\leq 2^n(a^{n+1}+b^{n+1})$, not some variation of that -- exactly that.
Discussion on question by F.V.: Prove…
Discussion on question by F.V.: Prove…