Analog of $(a+b)^2 \leq 2(a^2 + b^2)$ Is there an inequality such as
$$(a+b)^2 \leq 2(a^2 + b^2)$$
for higher powers of $k$
$$(a+b)^k \leq C(a^k + b^k)?$$
The generalized mean inequality states that
$$\dfrac{a+b}2\leq \left(\dfrac{a^k+b^k}2\right)^{1/k},$$
with equality if and only if $a=b$, from which it follows that
$$(a+b)^k\leq 2^{k-1}(a^k+b^k),$$
with equality if and only if $a=b$. Thus $C=2^{k-1}$ works and no smaller $C$ works.
The first inequality is true because : $2ab ≤ a^2+b^2$

There is no specific inequality for higher powers of k, although you can always give a very higher value for 'C' to make the second inequality true.