11.用数学归纳法。
n=2时不等式变为√(a1a2)+√(b1b2)<=√[(a+b1)(a2+b2)].①
平方得a1a2+b1b2+2√(a1a2b1b2）<=(a1+b1)(a2+b2),
整理得2√(a1a2b1b2)<=a1b2+a2b1,
由均值不等式知上式成立，命题成立；
假设n=k(k为大于1的整数)时命题成立，即
(∏ai)^(1/k)+(∏bi)^(1/k)<=[∏(ai+bi)]^(1/k),②那么
(∏ai)^[1/(k+1)]+(∏bi)^[1/(k+1)]
=√(∏ai)^[2/(k+1)]+√(∏bi)^[2/(k+1)]
<=√{(∏ai)^[2/(k+1)]+(∏bi)^[2/(k+1)]}
*√{a^[2/(k+1)]+b^[2/(k+1)]}(由①)③
<={(∏ai)^[1/(k+1)]+(∏bi)^[1/(k+1)]}
*{a^[1/(k+1)]+b^[1/(k+1)]}(平方得);
(∏ai)^[1/(k+1)]+(∏bi)^[1/(k+1)]
={(∏ai)^[k/(k+1)]}^(1/k)+{(∏bi)^[k/(k+1)]}^(1/k)
<=∏{ai^[k/(k+1)]+bi^[k/(k+1)]}^(1/k)(由②)
<=∏{ai^[1/(k+1)]+bi^[1/(k+1)]}(k次方得),
代入③即得命题成立。
