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

问题描述:

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即:a^2+aba^2+ab+c^2a^2-2ac+c^2∴(a-c)^2又∵c^2>ab
即c^2-ab>0
∴-√(c^2-ab)即c-√(c^2-ab)

(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+b0,b>0
∴a*a+a*b