√x+√y≤t√(x+y)恒成立等价于t≥(√x+√y)/√(x+y)恒成立,即t≥max{(√x+√y)/√(x+y)}
注意到[(√x+√y)/√(x+y)]²=(x+y+2√xy)/(x+y)=1+2√xy/(x+y)≤1+(x+y)/(x+y)=1+1=2
∴(√x+√y)/√(x+y)≤√2
且当x=y时(√x+√y)/√(x+y)=√2
于是max{(√x+√y)/√(x+y)}=√2
∴t≥√2,其取值范围为[√2,+∞)