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

2 Answers 2

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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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    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$2012-06-13
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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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    @ 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 result2012-06-13