Question

已知X,Y属于正实数,且X+Y+1/X+1/Y=5,则X+Y的最大值是

已知X,Y属于正实数,且X+Y+1/X+1/Y=5,则X+Y的最大值是

Answer 1

x+y+1/x+1/y
=(x+y)(1+1/(xy))
令x+y=k
4xy<=(x+y)²=k²
1+1/(xy)>=1+4/k²
所以(x+y)(1+1/(xy))>=k(1+4/k²)=k+4/k
也就是k+4/k<=5
解得1<=k<=4
k最小值是1，最大值是4
即x+y的最大值是4

Answer 2

依Cauchy不等式，得
5＝x+y+1²/x+1²/y≥x+y+(1+1)²/(x+y)
→(x+y)²-5(x+y)+4≤0
→1≤x+y≤4.
∴(x+y)|min＝1, (x+y)|max＝4。