Proof of Convexity for Power Function with Positive Exponent

[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 don't know
how to prove that $$f(t)=t^p$$ is negative
its pure parametric thing

??

 

[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.