Proof of U(pie(x)+(1-pie)y) > pie*U(x)+(1-pie)U(y) using sqrt property

[3.141592654]

Homework Statement

For U(w)=sqrt(w), prove that U(pie(x)+(1-pie)y) > pie*U(x)+(1-pie)U(y)

Homework Equations

sqrt(x)=x^(1/2)

The Attempt at a Solution

I have:

sqrt(pie(x)+(1-pie)y) > pie*sqrt(x)+(1-pie)sqrt(y) so...

(pie(x)+(1-pie)y)^(1/2) > pie*(x)^1/2+(1-pie)(y)^1/2

From here I don't know where to go. I don't have much experience with proofs so can anyone give some guidance? Thanks for your help!

 

[badphysicist]

Try squaring both sides.

 

[3.141592654]

alright, then (pi(x)+(1-pi)y) > pi^2*(x)+(1-pi)^2*(y). Therefore pi+(1-pi)>(pi^2)+((1-pi)^2)
=1 > (pi^2)+((1-pi)^2)
=1-(pi^2)>(1-pi)^2

Is this sufficient? What I don't quite get is that if you then square both sides, don't you get sqrt((1-(pi^2))) > sqrt((1-pi)^2)

=sqrt(1)-pi > (1-pi)
=1-pi > 1-pi ?? This obviously can't be right, so where did I go wrong? Thanks again.

 

[HallsofIvy]

Here "pi" is NOT the usual "pi", it is a number between 0 and 1, right? This is essentially a proof that the square root function is concave.

 

[3.141592654]

yes exactly, sorry for not making that clear before.