已知正数a，b，c满足a+b+c=10,a^2+b^2=c^2.则ab最大值为多少？

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

a+b=10-c,a^2+b^2=c^2
代入柯西不等式(a+b)^2<=2(a^2+b^2)得c^2+20c-100>=0
所以c>=10（√2-1)
ab=[(a+b)^2-(a^2+b^2)]/2=[(10-c)^2-c^2]/2=50-10c<=50-100(√2-1)
=150-100√2
所以ab最大值为150-100√2

-50 (-3 + 2 Sqrt[2])
a -> 10 -
5 (2 - Sqrt[2]) - (-50 + 50 (2 - Sqrt[2]) -
25 (2 - Sqrt[2])^2)/(-10 + 5 (2 - Sqrt[2]))
b -> 5 (2 - Sqrt[2])