1
$\begingroup$

How to prove for $s<1$$$|a+b|^s\le|a|^s+|b|^s$$ I tried to prove $|a+b|\le (|a|^s+|b|^s)^{\frac{1}{s}}$, but $\frac{1}{s}$ may not be integer, so I do not know how to expand it if it is not integer.

Is it the right way to prove it by trying to expand it using binomial expansion? How to prove it?

Thanks!

$\endgroup$
  • $\begingroup$ Where do $a$ and $b$ live? Also, do you mean $\left|b\right|^s$ in all instances? $\endgroup$ – Stahl Feb 18 '15 at 22:58
  • $\begingroup$ @Stahl Sorry, that is a typo. $\endgroup$ – Sherry Feb 18 '15 at 23:04
2
$\begingroup$

for positive $a,b$ and $s \in (0,1)$ $$ a^s +b^s = s\left(\int_0^a x^{s-1}dx+\int_0^b x^{s-1}dx \right) \\ \ge s\left(\int_0^a x^{s-1}dx+\int_0^b (x+a)^{s-1}dx \right) \\ = s\left(\int_0^a x^{s-1}dx+\int_a^{a+b} x^{s-1}dx \right) \\ = s\left(\int_0^{a+b}x^{s-1}dx \right) \\ = (a+b)^s $$

$\endgroup$
1
$\begingroup$

Hint: Consider the function $f(x)=(1+x)^s-1-x^s$ for $x\geq 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.