# Proof question 2,1

1. Oct 26, 2009

### lom

it is given that
$$1\leq p< +\infty\\$$
$$\alpha ,\beta >0 \\$$
$$a,b\geq 0\\$$

prove that
$$(\alpha a+\beta b )^p\leq (\alpha +\beta )^p\left ( \frac{\alpha }{\alpha +\beta }a^p+\frac{\beta }{\alpha +\beta }b^p \right )$$

hint: prove first that $$f(t)=t^p$$ is a convex
on this region $$[0,+\infty)$$
reminder: function f(t) is called convex on some region if for every b,a
and on
$$0\leq \lambda \leq 1\\$$
we have
$$f(\lambda a +(1-\lambda)b)\leq\lambda f(a)+(1-\lambda)f(b)$$

my thoughts:
i know from calc1 that a function is convex if its second derivative is negative or something (i am not sure)

i dont know
how to prove that $$f(t)=t^p$$ is negative
its pure parametric thing

??

2. Oct 26, 2009

### Dick

You've got it backwards. f(t) is convex if the second derivative is positive, not negative. So you don't want to show f(t) is negative, you want to show f''(t) is positive.

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