Question

(1)已知a,b是正数,且a+b=1.求证:(ax+by)(ay+bx)≥xy.

(2)若x+3y-1=0,求证:2^x+8^y≥.

Answer 1

证明:(1)左边=(a²+b²)xy+ab(x²+y²)

=［(a+b)²-2ab］xy+ab(x²+y²)

=(1-2ab)xy+ab(x²+y²)

=xy+ab(x-y)²,

∵a＞0,b＞0,(x-y)²≥0,

∴左边≥xy=右边.

因此不等式成立.

(2)2^x+2^3y≥.