Possible Duplicate:
Showing the inequality $|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$
In the condition $a,b \in[0,\infty)$, $1\le p<\infty$,
How can I conclude this inequality? $$(a+b)^p \le 2^{p-1} (a^p + b^p)$$
Asked
Active
Viewed 529 times
1
2 Answers
1
you could solve it using Hölder inequality. $ (a+b)^p \leq 2^{p-1}(a^p+b^p) \Leftrightarrow (a+b) \leq (1+1)^{1- \frac{1}{p}} (a^p +b^p)^{\frac{1}{p}} $ and then apply Hölder.
-
@ ckark : I prefer that the $\Rightarrow$ be replaced by $\Leftarrow$. Ins'nt it ? – Mohamed Jun 13 '12 at 06:12
-
I'm not familiar with that inequality but it seems to be very useful. – wowhapjs Jun 13 '12 at 06:18
-
way right! It could be replaced with a $ \Rightleftarrow $ also :D – clark Jun 13 '12 at 06:31
-
hmm propably better $\Leftarrow$ but I will not change it again – clark Jun 13 '12 at 06:32
-
sorry I rushed before, the inequality is $\sum_{i=0}^{n}a_i b_i \leq =(\sum_{i=0}^{n}a_i^q )^{\frac{1}{q}}(\sum_{i=0}^{n} b_i^p)^{\frac{1}{p}}$ where $\frac{1}{q} + \frac{1}{p}=1$ – clark Jun 13 '12 at 06:38
0
Hint: $f(x)=x^p$ and convexity
$f(\frac a2 + \frac b2) \leq \frac 12 f(a) + \frac 12 f(b)$
Mohamed
- 3,771
-
-
@ wowhapjs You hav'nt $p$ an integer. You replace $f\left(\frac{a+b}{2} \right)$ by it's value : $\left(\frac{a+b}{2}\right)^p$ and the same for $f(a)=a^p$ and $f(b)=b^p$ and you have the result – Mohamed Jun 13 '12 at 07:32