- #1
brntspawn
- 12
- 0
For 0<x<y, show that x<[tex]\sqrt{xy}[/tex]<1/2(x+y)<y
I have no difficulty showing that x<[tex]\sqrt{xy}[/tex] and 1/2(x+y)<y. I am having difficulty with [tex]\sqrt{xy}[/tex]<1/2(x+y).
x<y
xx<xy
x[tex]^{2}[/tex]<xy
x<[tex]\sqrt{xy}[/tex]
and
x<y
x+y<y+y
x+y<2y
1/2(x+y)<y
I have no difficulty showing that x<[tex]\sqrt{xy}[/tex] and 1/2(x+y)<y. I am having difficulty with [tex]\sqrt{xy}[/tex]<1/2(x+y).
x<y
xx<xy
x[tex]^{2}[/tex]<xy
x<[tex]\sqrt{xy}[/tex]
and
x<y
x+y<y+y
x+y<2y
1/2(x+y)<y