Prove (a+b) p

That is 2 (a+b) p.

Direct evidence [(a+b)/2] p

Constructor f (x) = xp (x >; 0)

f''(x)=p(p- 1)x^(p-2)>； 0 so f(x) is convex.

From convexity (that is, piano inequality), we can know [f (a)+f (b)]/2 >; =f[(a+b)/2]

It's [(a+b)/2] p.