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Prove that $(a + b)^p \leq a^p + b^p$ for all $a, b \geq 0$ and $0 < p < 1$.

This was found in a section of Royden's Analysis on metric spaces, but the result does not appear to have anything to do with metric spaces.

My initial strategy was to show that $a^p + b^p - (a + b)^p \geq 0$, but that was not fruitful.

Taking the natural log of things doesn't seem helpful either:

$\ln((a + b)^p) = p\ln(a + b)$

My next idea was that, since norms were defined in this section, perhaps $\|u\| = u^p$ defines a norm, in which case the triangle inequality would hold. Well, it's not a norm because $\|\alpha{}u\| = \alpha\|u\|$ does not hold, but besides, the triangle inequality is what I wanted to show anyways.

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2 Answers 2

up vote 4 down vote accepted

If $a = b = 0$, the inequality trivially holds. So let's suppose $a+b > 0$. Then we can divide by $(a+b)^p$, and the inequality to show becomes

$$1 \leqslant \left(\frac{a}{a+b}\right)^p + \left(\frac{b}{a+b}\right)^p.$$

That is relatively easy to see, since for $0 \leqslant x \leqslant 1$ and $0 < p < 1$ we have $x \leqslant x^p$.

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"Relatively easy to see" is not really an answer is it? –  Myself Jan 27 at 22:37
    
Okay, let's make it more obvious then. –  Daniel Fischer Jan 27 at 22:39
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Here are two proofs:

1st: For any $t>0$, we have $t^{p-1}>(a+t)^{p-1}$. Then integrate both sides from 0 to b.

2nd: We know that $l_p$ spaces($L_p$ spaces with counting measure) are nested as follows: if $0<p<q\le\infty$, then $||f||_q\le||f||_p$. With this in mind, you can see that $(a+b)^p$ is $p$-th power of 1-norm of a function defined on two points with values $a$ and $b$. So it must be no greater than $p$-th power of the $p$-norm, which is $a^p+b^p$.

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Thank you for sharing the second proof especially, because it shows how this seemingly useless inequality fits into Real Analysis. –  user2521987 Jan 27 at 23:18
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