For \(a,b\ge 0\) and \(k\in \mathbb{N}\), we have
\[ (a+b)^k\le 2^k(a^k+b^k). \]
Proof
Proof by induction. Using the induction hypothesis, we deduce
\[ (a+b)^{k+1}\le 2^k(a^k+b^k)(a+b)=2^k(a^{k+1}+b^{k+1}+ab^{k}+a^kb). \]Then, by Young’s inequality for products with \(p=(k+1)\) and \(q=\frac{k+1}{k}\),
\[ ab^k\le \frac{1}{k+1}a^{k+1}+\frac{k}{k+1}b^{k+1}, \]and similarly
\[ a^kb\le \frac{k}{k+1}a^{k+1} + \frac{1}{k+1}b^{k+1}. \]Combining all together, it follows that
\[ (a+b)^{k+1}\le 2^{k+1}(a^{k+1}+b^{k+1}). \]