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.