a,b,c,d都为正数,a+b=c+d 且c>a>b>d,怎么证明ab>cd?

设a+b=c+d=k,则
a=k-b
c=k-d
ab=(k-b)b=kb-b²
cd=(k-d)d=kd-d²
ab-cd=(kb-b²)-(kd-d²)=kb-b²-kd+d²=(kb-kd)-(b²-d²)=k(b-d)-(b+d)(b-d)=(b-d)[k-(b+d)]
∵a+b=c+d=k,
∴k=(a+b+c+d)/2
∴ab-cd=(b-d)[k-(b+d)]=(b-d)[(a+b+c+d)/2-(b+d)]=(b-d)[(a+c-b-d)/2]=(b-d)[(a-b)+(c-d)]/2
∵c>a>b>d
∴b-d>0,a-b>0,c-d>0
∴ab-cd=(b-d)[(a-b)+(c-d)]/2>0
∴ab>cd