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