應(yīng)該是求證：［a(1)＋a(2)＋a(3)＋······＋a(n)］［1/a(1)＋1/a(2)＋1/a(3)＋······＋1a(n)］≧n^2.
［證明］
構(gòu)造二次函數(shù)：y＝（√kx＋1/√k）^2＝kx^2＋2x＋1/k,其中k是正數(shù).
顯然有：kx^2＋2x＋1/k≧0.
依次令k＝a(1)、a(2)、a(3)、a(4)、······、a(n),得：
a(1)x^2＋2x＋1/a(1)≧0,
a(2)x^2＋2x＋1/a(2)≧0,
a(3)x^2＋2x＋1/a(3)≧0,
······
a(n)x^2＋2x＋1/a(n)≧0.
將以上n個不等式左右分別相加,得：
［a(1)＋a(2)＋a(3)＋······＋a(n)］x^2＋2nx＋［1/a(1)＋1/a(2)＋1/a(3)＋······＋1a(n)］≧0.
令f（x）＝［a(1)＋a(2)＋a(3)＋······＋a(n)］x^2＋2nx＋［1/a(1)＋1/a(2)＋1/a(3)＋······＋1a(n)］
∵k＞0,∴a(1)＋a(2)＋a(3)＋······＋a(n)＞0,
∴f（x）是一條開口向上的拋物線,
∴要滿足f（x）≧0,就需要：
（2n）^2－4［a(1)＋a(2)＋a(3)＋······＋a(n)］［1/a(1)＋1/a(2)＋1/a(3)＋······＋1a(n)］≦0,
∴［a(1)＋a(2)＋a(3)＋······＋a(n)］［1/a(1)＋1/a(2)＋1/a(3)＋······＋1a(n)］≧n^2.
注：括號“( )”里的數(shù)字是下標.