Prove that $ (a+b)^{2n} \leq 2^{2n-1}(a^{2n}+b^{2n})$

Question

I have to prove that for all $(a;b) \in \mathbb{R}^2$, and for all $n \in \mathbb{N}$ we have: $$ (a+b)^{2n} \leq 2^{2n-1}(a^{2n}+b^{2n})$$ without using induction.

I tried to use the convexity of $x^{2n}$ Which gave me that $x^{2n} \geq 2nx+1-2n$ but I didn’t succeed.

Answer 1

If $a,b \geqslant 0$ then, by the power mean inequality

$$M_{2n} \geqslant M_1$$ $$\left(\frac{a^{2n}+b^{2n}}{2}\right)^{\frac{1}{2n}} \geqslant \frac{a+b}{2} $$ and the result follows. If one of $a$ and $b$ is negative, then the LHS (in the question) reduces but the RHS remains unchanged. If both are negative then neither side changes. And so we are done.

Answer 2

Here's an alternative solution for the case $a$, $b\ge0$. When $n=0$, it's $1\le1$.

When $n\in\Bbb N_+$, $1^{\frac{2n}{2n-1}}=1$, Holder inequality gives $$a+b\le(1+1)^{\frac{2n-1}{2n}}\cdot\left(a^{2n}+b^{2n}\right)^{\frac1{2n}}.$$ Therefore $(a+b)^{2n}\le2^{2n-1}(a^{2n}+b^{2n})$, which is the desired inequality

Answer 3

Let $f(x)=x^{2n}$.

Since $f$ is a convex function (for example, $f''(x)=2n(2n-1)x^{2n-2}\geq0$), by Jensen we obtain: $$\frac{a^{2n}+b^{2n}}{2}\geq\left(\frac{a+b}{2}\right)^{2n},$$ which is your inequality.