设a>0,b>0,2c>a+b,求证:c-根号下(c^2 - ab)

设a>0,b>0,2c>a+b,求证:c-根号下(c^2 - ab)
设a>0,b>0,2c>a+b,求证:c-根号下(c^2 - ab)

设a>0,b>0,2c>a+b,求证:c-根号下(c^2 - ab)
证明:c-√(c^2-ab)

(1) 4c^2>a^2+b^2+2ab≥2ab+2ab≥4ab
c^2>ab
(2)∵a+b<2c, 且a>0,b>0
∴a*a+a*b即：a^2+ab<2ac
a^2+ab+c^2<2ac+c^2
a^2-2ac+c^2∴(a-c)^2∴-√(c^2-ab)即c-√(c^2-ab)