设x>-1,求函数y=(x^2+7x+10)/(x+1)的最值?若x、y是正数,则(x+1/2y)^2+(y+1/2x)^2的最小值?

Question

设x>-1,求函数y=(x^2+7x+10)/(x+1)的最值?若x、y是正数,则(x+1/2y)^2+(y+1/2x)^2的最小值?

Answer 1

(x^2+7x+10)/(x+1)=5+(x+1)+[4/(x+1)]≥5+2√[(x+1)*4/(x+1)]=9
最小值9
(x+1/2y)^2+(y+1/2x)^2=x^2+(1/4x^2)+(x/y)+(y/x)+y^2+(1/4y^2)
≥2*[x*1/2x]+2√[(x/y)*(y/x)]+2*[y*1/2y]=4
最小值4