正数A，B，C满足A+B+C=10，且A^2+B^2=c^2。则AB的最大值为？

a+b+c=10; a^2+b^2=c^2所以
(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc
=2c^2+2ab+2c(a+b)
=2c^2+2ab+2c(10-c)
=2ab+20c=100;
所以
ab=50-10c
=50-10(10-a-b)
=10(a+b)-50
>=20根号(ab)-50
ab-20根号(ab)+50>=0;
解得 根号(ab)<=10-5根号2
所以 ab<=(10-5根号2)^2=150-100根号2;
即ab的最大值为150-100根号2