Question

求分式的最值x>0,y>0,求(x^2+y^2+1)/(x+2*y)的最小值,

Answer 1

利用Cauchy不等式和基本不等式得,

(x²+y²+1)/(x+2y)

＝[x²/1+(2y)²/4+1]/(x+2y)

≥[(x+2y)²/(1+4)+1]/(x+2y)

＝(x+2y)/5+1/(x+2y)

≥2√[(x+2y)/5·1/(x+2y)]

＝(2√5)/5.

故所求最小值为：(2√5)/5.

且最小值时,有

x:1＝y:2,和(x+2y)/5＝1/(x+2y),

解得,x＝√5/5,y＝2√5/5.