又(a^2b b^2a)/("(" ab")" ^(a+b) )=ab-a·ba-b=(b/a)^(a"-" b),

　　当a=b时,(b/a)^(a"-" b)=10=1;

　　当a>b>0时,00,所以(b/a)^(a"-" b)<1;

　　当b>a>0时,b/a>1,a-b<0,所以(b/a)^(a"-" b)<1.

　　所以(a^2b b^2a)/("(" ab")" ^(a+b) )=(b/a)^(a"-" b)≤1.

　　综上可知a2bb2a≤(ab)a+b.

10.导学号26394032已知θ∈(π/4 "," π/2),且a=cos 2θ,b=cos θ-sin θ,试比较a与b的大小.

解因为θ∈(π/4 "," π/2),所以2θ∈(π/2 "," π).

　　所以a=cos 2θ<0,且cos θ

　　因为a/b=cos2θ/(cosθ"-" sinθ)=(cos^2 θ"-" sin^2 θ)/(cosθ"-" sinθ)

　　=cos θ+sin θ=√2sin(θ+π/4),

　　又θ∈(π/4 "," π/2),所以θ+π/4∈(π/2 "," 3π/4),

　　所以sin(θ+π/4)∈(√2/2 "," 1).

　　所以√2sin(θ+π/4)∈(1,√2).即a/b>1,故a