已知a>0,b>0,2c>a+b求证（1）c^2>ab (2)c-√c^2-ab

已知a>0,b>0,2c>a+b求证（1）c^2>ab (2)c-√c^2-ab

(1) 4c^2>a^2+b^2+2ab≥2ab+2ab≥4ab
c^2>ab
(2)∵a+b0,b>0
∴a*a+a*b

(1) 4c^2>a^2+b^2+2ab>2ab+2ab=4ab
c^2>ab
(2)看不懂，把括号写出来啊

(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又∵c^2>ab
即c^2-ab>0
∴-√(c^2-ab)即c-√(c^2-ab)