已知a>b>c,求證：1/(a-b)+1/(b-c)≥4/(a-c)

熱心網(wǎng)友

左邊通分得:(a-c)/(a-b)(b-c),所以只需證(a-c)/(a-b)(b-c)≥4/(a-c)即證(a-c)^2≥4(a-b)(b-c),即證:(a-b)(b-c)≤(a-c)^/4因?yàn)閤y≤[(x+y)/2]^2,將a-b看成是x,將b-c看成是y則(a-b)(b-c)≤{[(a-b)+(b-c)]/2}^2=[(a-c)/2]^2=(a-c)^2/4得證

熱心網(wǎng)友

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