Sum of numbers in $[0,1]$ raised to exponents at least as large as $1$.

Question

For $a,b\in [0,1]$ and $\epsilon\geq 0$, does the following equality hold?

$a^{1+\epsilon}+b^{1+\epsilon}\geq |a-b|^{1+\epsilon}$

All I can think to do so far is: \begin{align*} |a-b|^{1+\epsilon} = ((a-b)^2)^{(\epsilon+1)/2} \end{align*}

I want to know if I can apply this result, I know such a result is true for $\epsilon=1$, i.e. $a^2+b^2\geq (a-b)^2$

Answer 1

Without loss of generality consider $a\ge b \ge 0$, then

$$a^{1+\epsilon} + b^{1+\epsilon} \ge a^{1+\epsilon} \ge \lvert a-b\rvert^{1+\epsilon}.$$

Using the increasing property of $x^{1+\epsilon}$ for $x\ge 0$:

$$\begin{align*} b &\ge 0 &&\implies & b^{1+\epsilon} &\ge 0\\ a &\ge a-b \ge 0 &&\implies & a^{1+\epsilon} &\ge (a-b)^{1+\epsilon} \end{align*}$$

Answer 2

As ahint :
Without lose of generality divide by $b^{1+\epsilon}$ so $$a^{1+\epsilon}+b^{1+\epsilon}\geq |a-b|^{1+\epsilon} \ \div b^{1+\epsilon}\\ (\frac ab)^{1+\epsilon}+1\ge |(\frac ab)-1|^{1+\epsilon}$$now take $\frac ab=x $ and $x \in [0,\infty)$
now take $f(x)=x^{1+\epsilon}+1-|x-1|^{1+\epsilon}$
note that $|x-1|^{1+\epsilon}=|1-x|^{1+\epsilon}$
now try to prove $f(x)\ge 0$
I write for the case of $0\leq x<1$ and rest for you
$$0\leq x<1 \to f(x)=x^{1+\epsilon}+1-(1-x)^{1+\epsilon}\\f'(x)=(1+\epsilon)x^{\epsilon}-(1+\epsilon)(1-x)^{\epsilon}(-1)=\\(1+\epsilon)(x^{\epsilon}+(1-x)^{\epsilon})\ge 0$$so $f(x)$ is increasing function, and range can be obtained by check $f(0),f(\frac 12),\lim_{x\to \infty} f(x)$
Remark $x=\frac 12 $ is the roof of $f'(x)=0$
$$f(0)=0\\f(\frac 12)=1\\\lim_{x\to \infty} f(x)\to +\infty \\so\\ f(x) \in [0,+\infty) \to 0\leq f(x) <\infty$$ and inequality proved for the case of $0\leq x<1$
I think you can take over