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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? $$(a+b)^p \le 2^{p-1} (a^p + b^p)$$

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marked as duplicate by user17762, Henry, Phira, t.b., Zev Chonoles Jun 14 '12 at 4:12

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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.

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@ ckark : I prefer that the $\Rightarrow$ be replaced by $\Leftarrow$. Ins'nt it ? – Mohamed Jun 13 '12 at 6:12
I'm not familiar with that inequality but it seems to be very useful. – wowhapjs Jun 13 '12 at 6:18
way right! It could be replaced with a $ \Rightleftarrow $ also :D – clark Jun 13 '12 at 6:31
hmm propably better $\Leftarrow$ but I will not change it again – clark Jun 13 '12 at 6:32
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 6:38
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Hint: $f(x)=x^p$ and convexity

$f(\frac a2 + \frac b2) \leq \frac 12 f(a) + \frac 12 f(b)$

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Then can I conclude inequality by induction? – wowhapjs Jun 13 '12 at 6:16
@ 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 7:32

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