solving polynomial equation (showing non-negativity)

Question

I would like to show that for all $a,b\geq 0$, and integer $k\geq 1$, we have $(a+b)^k\leq 2^k(a^k+b^k)$. Is there an elegant way to prove this other than induction?

Thanks

Answer 1

This is equivalent as $$\left(\frac{a+b}{2}\right)^k\leq a^k+b^k$$ Which is clearly true since $f(x)=x^k$ is strictly increasing, and $$\min(a,b)\leq \frac{a+b}{2}\leq\max(a,b).$$

To explain further, suppose $a\leq b$, then because $f$ is increasing you have $$f(a)\leq f\left(\frac{a+b}{2}\right)\leq f(b) \leq f(a)+f(b).$$

Thus it even holds if you replace the rhs of your original inequality by $2^k\max(a^k,b^k)$.

Answer 2

Note that $$ a+b\leq2\max(a,b)\implies(a+b)^k\leq2^k(\max(a,b))^k=2^k(\max(a^k,b^k))\leq2^k(a^k+b^k). $$ where we used $x^k$ is strictly increasing for $x\ge 0$.