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I believe the following inequality is true. Can anyone prove it?

Let $a,b \in \mathbb{R}$ with $a,b \neq 0$ and $0 \leq \alpha < 1$, then

$$ \left| \dfrac{a}{|a|^{\alpha}} - \dfrac{b}{|b|^{\alpha}} \right| \leq 2^{\alpha}|a - b|^{1-\alpha}. $$

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If you allow $a=0$, $\dfrac{a}{|a|^\alpha}$ won't make sense. – user21436 Jan 31 '12 at 19:34
Yes, of course. I believe they should be nonzero. – al0 Jan 31 '12 at 19:39
Have you tried something/met some dead ends? – Ravi Jan 31 '12 at 19:45
up vote 5 down vote accepted

There are two cases to consider, depending on whether $a$ and $b$ have the same or opposite sign. If $a$ and $b$ have the same sign, after dividing by $|a|^{1-\alpha}$ the inequality says $|1 - t^{1-\alpha}| \le 2^\alpha |1-t|^{1-\alpha}$. We can assume wlog $0 < t \le 1$ (otherwise interchange $a$ and $b$), so it becomes $t^{1-\alpha} + 2^\alpha (1-t)^{1-\alpha} \ge 1$. This is true at $t=0$ and at $t=1$, and the right side is concave as a function of $t$ so it is true for $0 \le t \le 1$ (and you could replace the $2^\alpha$ by $1$).

If $a$ and $b$ have opposite sign, the inequality becomes $1 + t^{1-\alpha} \le 2^\alpha (1+t)^{1-\alpha}$ for $t > 0$. Taking $s = t^{1-\alpha}$ and $p = 1/(1-\alpha)$, it says $1 + s \le 2^{1-1/p} (1 + s^p)^{1/p}$ for $s > 0$ and $p>1$. This is true because the right side is a convex function of $s$, and the left side is the equation of its tangent line at $s=1$.

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The following assumes that $a$ and $b$ are positive. Edit: Later, we deal with the other sign possibilities. The proofs use basic calculus.

Without loss of generality, we may assume that $a\ge b$. The inequality is homogeneous, so we only need to show that if $x\ge 1$ then $$x^{1-\alpha}-1 \le 2^\alpha (x-1)^{1-\alpha}.$$

Let $f(x)=2^\alpha(x-1)^{1-\alpha}-(x^{1-\alpha} -1)$. We have $f(1)\ge 0$. A straightforward calculation shows that $f'(x)>0$ for all $x\ge 1$. For note that $$f'(x)=(1-\alpha)(2^\alpha)(x-1)^{-\alpha}-(1-\alpha)x^{-\alpha},$$ so we need only check that $2^\alpha x^\alpha\ge (x-1)^\alpha$, or equivalently that $2x>x-1$.

Added: If $a$ and $b$ are both negative, the same idea works, since the ratio is positive. To deal with the remaining cases, we may assume that the number of largest absolute value is positive. So now we want to show that if $x\ge 1$, then $x^{1-\alpha}+1 \le 2^{\alpha}(x+1)^\alpha$.

Let $g(x)=2^\alpha(x+1)^{1-\alpha} -(x^{1-\alpha} +1)$. We have $g(1)=0$. We show that $g'(x)>0$ if $x>1$. By a short calculation, this is equivalent to showing that $2^\alpha x^\alpha> (x+1)^\alpha$. This is true if $2x>x+1$, that is, if $x>1$.

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