Question

已知a>b>0,求证√ a²-b² + √ 2ab-b² >a

Answer 1

右边的式子可以写作a^(b/(a
b))*b^(a/(a
b))
左除以右：a^[1/2-b/(a
b)]*b^[1/2-a/(a
b)]
=a^[1/2(a-b)/(a
b)]*b^[1/2(b-a)(a
b)
=(a/b)^[1/2(a-b)(a
b)]
现在判断：当a
b时，a/b大于1，指数大于0，原式大于1；当a=b时原式等于1，所以原不等式成立