Prove that $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ by induction. 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 $a,b \geq 0$, $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ holds.
IS: For $n \rightarrow n + 1$ we get.
\begin{align*}
    &\hspace{0.55cm} (a+b)^{n+1} \leq 2^n(a^{n+1} + b^{n+1}) \\
    &= (a+b)^n  (a+b) \leq 2^n(a^na + b^nb) \\
    &= (a+b)^n  (a+b) \leq 2^n(a^n+b^n)  (a+b) \\
    &= (a+b)^n \leq 2^n(a^n+b^n) \\
    &= \frac{(a+b)^n}{2} \leq 2^{n-1}(a^n+b^n) \\
    &\rightarrow \frac{(a+b)^n}{2} \leq (a+b)^n \leq 2^{n-1}(a^n+b^n)
\end{align*}
Because of IH $(a+b)^n \leq 2^{n-1}(a^n+b^n)$, $0.5(a+b)^n \leq 2^{n-1}(a^n+b^n)$ must also hold for $a,b \geq 0$ and $n \in \mathbb{N}$. Thus the formula holds for all $n$
Is this a valid proof? I am not sure if the last part is correct and if I can simply divide by $(a+b)^n$, because it can be zero
2nd try:
\begin{align*}
    &\hspace{0.95cm} (a+b)^{n+1} = (a+b)^n(a+b)\\
    &\stackrel{\text{IV}}{\longrightarrow} (a+b)^n(a+b) \leq 2^{n-1}(a^n + b^n)(a+b) \\
    &\Longleftrightarrow 2{(a+b)^{n+1}} \leq 2^n(a^n+b^n)  (a+b) \\
    &\Longleftrightarrow  2{(a+b)^{n+1}} \leq 2^n(a^na+b^n*b + a^nb + b^na) \\
    &\Longleftrightarrow  2{(a+b)^{n+1}} \leq 2^n(a^{n+1}+b^{n+1} + a^nb + b^na) \\
    &\longrightarrow  (a+b)^{n+1} < 2{(a+b)^{n+1}} \leq 2^n(a^{n+1}+b^{n+1} + a^nb + b^na) \\
    &\longrightarrow (a+b)^{n+1} \leq 2^n(a^{n+1}+b^{n+1} + a^nb + b^na)
\end{align*}
It remains to show that $a^{n+1} + b^{n+1} \geq a^nb + b^na$ holds, because then the statement also holds for $n+1$ $\rightarrow$ $(a+b)^{n+1} \leq 2^n(a^{n+1}+b^{n+1})$ and thus all $n \in \mathbb{N}$. It holds from above
\begin{align*}
    \hspace{1cm} a^{n+1} + b^{n+1} &\geq a^nb + b^na \\
    \Longleftrightarrow a^{n+1} + b^{n+1} - a^nb - b^na &\geq 0 \\
    \Longleftrightarrow a^n(a-b) - b^n(a - b) &\geq 0 \\
    \Longleftrightarrow (a^n-b^n)(a-b) &\geq 0 \\
\end{align*}
$(a^n-b^n)(a-b) \geq 0$ is satisfied for all $n \in \mathbb{N}$ and $a,b \geq 0$ and thus the inequality from the problem also holds for all $n$
 A: Unfortunately, your second attempt doesn't quite get you there. There are a few algebraic mistake, as well as, in general, $x/2\leq y$ does not mean that $x\leq y$. We will need the following lemma:

Lemma:  $a^nb+ab^n\leq a^{n+1}+b^{n+1}$ for all $a,b\geq0$ and $n\in\mathbb N$.
Proof. WLOG, assume $a\geq b$, and note that this implies $a^n\geq b^n$. We have
\begin{align*}
a-b\geq0\text{ and } a^n-b^n\geq 0&\implies(a^n-b^n)(a-b)\geq0\\
&\implies a^{n+1}-a^nb-ab^n+b^{n+1}\geq 0\\
&\implies a^{n+1}+b^{n+1}\geq a^nb+ab^n.
\end{align*}

To prove the main result, we use your proof and begin the induction:
\begin{align*}
(a+b)^{n+1}&=(a+b)^n(a+b)\\
&\leq 2^{n-1}(a^n+b^n)(a+b)&\text{by ind. hyp.}\\
&=2^{n-1}(a^{n+1}+b^{n+1}+a^nb+ab^n)\\
&=2^{n-1}(a^{n+1}+b^{n+1})+2^{n-1}(a^nb+ab^n)\\
&\leq2^{n-1}(a^{n+1}+b^{n+1})+2^{n-1}(a^{n+1}+b^{n+1})\\
&=2^{n}(a^{n+1}+b^{n+1}).
\end{align*}
A: Another solution which avoids induction is using Jensen's inequality. I assume from your attempt that $a, b \geq 0$.
Let $f(x) = x^n$ for some $n \in \mathbb{N}$. Note that $f$ is a convex function over the interval $(0, \infty)$. Then by Jensen we have
$$
\frac{(a+b)^n}{2^n} = f\bigg(\frac{a+b}{2}\bigg) \leq \frac{f(a)+f(b)}{2} = \frac{a^n+b^n}{2} \\
\iff (a+b)^n \leq 2^{n-1}\big(a^n+b^n\big).
$$
