$(1+x)^d \leq 1 + x^d$ for $x \geq 0$ and $d \in (0,1]$

Question

I come across the following inequality: for all $x \geq 0$ and $d \geq 0$,

• if $d \leq 1$:

$$(1+x)^d \leq 1 + x^d,$$

• if $d \geq 1$:

$$(1+x)^d \geq 1 + x^d.$$

I think they are related to convex functions and Jensen's inequality. But I have trouble proving them. Can anyone share some ideas?

Answer 1

Consider $$ f(x)=(1+x)^d-x^d, $$ then $$ f'(x)=d((1+x)^{d-1}-x^{d-1}). $$

Note that if $d\in(1,+\infty)$ ($d\in(0,1]$), then $f'(x)\geq 0$ for all $x$ ($f'(x)\leq 0$ for all $x$), so $f$ is monotonically increasing (decreasing). Can you end now?

Answer 2

When $d\le 1$, $f(t)=t^d$ is concave. Therefore, the slope $g(t)=f(t+1)-f(t)$ is non-increasing. Thus $(1+x)^d-x^d=g(x)\le g(0)=1$. Using a similar method, we can prove the case $d\ge 1$.

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To see why $g(t)$ is non-increasing, notice that $$g(t)=\int_{t}^{t+1}f'(u)du$$ and $f'(u)$ is non-increasing in $u$ due to concavity when $d\le 1$.