已知x>y>0，且xy=1，求證(x2+y2)/(x-y)>=2*2^1/2，并說明等號成立的條件

熱心網友

證:(x平方+y平方)/(x-y)=[(x-y)平方+2xy]/(x-y)=(x-y)+2xy/(x-y) =(x-y)+2/(x-y)≥2√(x-y)*2/(x-y)=2√2 當且僅當x-y=√2,時等號成立,由于xy=1,可知........x=(√6+√2)/2,y=(√6-√2)/2時等號成立

熱心網友

配項： (x^2 + y^2 - 2 * x * y + 2 * x * y) / (x - y) = 2 * 2^1/2又因為： x * y = 1所以： (x^2 + y^2 - 2 * x * y + 2) / (x - y) = 2 * 2^1/2 ((x - y)^2 + 2) / (x - y) = 2 * 2^1/2因為xy，左右同乘以(x - y): (x - y)^2 - 2 * 2^1/2 * (x - y) + 2 = 0于是： ((x - y) - 2^1/2)^2 = 0左邊是完全平方項，一定非負。當 x - y - 2^1/2 = 0 時即 x = y + 2^1/2 時取等號。好像一點都不難。

熱心網友

(x2+y2)/(x-y)=(x2+y2-2xy)/(x-y)+2/(x-y)==x-y+2/(x-y)≥2√[(x-y)*2/(x-y)]=2√2.等號成立x-y=2/(x-y)x=(√3+1)/√2,y=(√3-1)/√2.

熱心網友

已知xy0，且xy=1，求證(x^2+y^2)/(x-y)≥2*√2　，并說明等號成立的條件 (x^2+y^2)/(x-y) = [(x-y)^2 +2]/(x-y) = (x-y) + 2/(x-y)由已知條件及均值不等式得：(x-y) + 2/(x-y)≥2*√2當且僅當x-y = 2/(x-y)時，即 x-y =√2時，取等號由于xy=1 ,解得：x=(√6+√2)/2 、y=(√6-√2)/2