设a=x+m,d=x-m,b=x+n,c=x-n,x>m>n>0
(√a+√d)²=(x+m)+(x-m)+2√(x²-m²)=2x+2√(x²-m²)
(√b+√c)²=(x+n)+(x-n)+2√(x²-n²)=2x+2√(x²-n²)
因为x>m>n>0,于是x²>m²>n²
于是0<x²-m²<x²-n²
于是2x+√(x²-m²)<2x+√(x²-n²)
即(√a+√d)²<(√b+√c)²
得√a+√d<√b+√c
设a=x+m,d=x-m,b=x+n,c=x-n,x>m>n>0
(√a+√d)²=(x+m)+(x-m)+2√(x²-m²)=2x+2√(x²-m²)
(√b+√c)²=(x+n)+(x-n)+2√(x²-n²)=2x+2√(x²-n²)
因为x>m>n>0,于是x²>m²>n²
于是0<x²-m²<x²-n²
于是2x+√(x²-m²)<2x+√(x²-n²)
即(√a+√d)²<(√b+√c)²
得√a+√d<√b+√c