Is there a way to prove the next inequality without using the binomial series?

Question

I have $\forall a\ge 1, x>0$ and I want to show that: $$(x^a+1)^{1/a} \le (1+x)$$

The only way I see is through the binomial series.

Can you suggest other approaches?

Answer 1

Let $f(x)=(1+x)^{a}-x^{a}$ for $x \geq 0$. Then $f'(x)=a(1+x)^{a-1}-ax^{a-1} \geq 0$. Also $f(0)=1$. Hence $f(x) \geq 1$ for all $x>0$. This gives $(1+x)^{a}-x^{a}\geq 1$ which is equivalent to the given inequality.

Answer 2

Let $f(x)=x^a$.

Thus, $f$ is a convex function.

Thus, since $(x+1,0)\succ(x,1)$ or $(x+1,0)\succ(1,x)$, by Karamata we obtain: $$f(x+1)+f(0)\geq f(x)+f(1) $$ or $$(x+1)^a+0\geq x^a+1$$ and we are done!