If x > 1, it is proved that LNX > (2 (x-1)) / (x + 1)

If x > 1, it is proved that LNX > (2 (x-1)) / (x + 1)

F (x) = lnx-2 (x-1) / (x + 1), f '(x) = 1 / X - [2 (x + 1) - 2 (x-1)] / (x + 1) ^ 2 = 1 / x-4 / (x + 1) ^ 2 = [(x + 1) ^ 2-4x] / [x (x + 1) ^ 2] = (x-1) ^ 2 / [x (x + 1) ^ 2] > 0, when x > 1, and f (1) = 0, then f (x) > F (1) = 0, that is, LNX > 2 (x-1) / (x + 1)

Given the function f (x) = lnx-x + 1, prove LN2 ^ 2 / 2 ^ 2 + Ln3 ^ 2 / 3 ^ 2 + +lnn^2/n^2=2)

It is proved that P = 1
f(x)=lnx-x+1,x>=1
f'(x)=(1-x)/x1
Then f (x) decreases monotonically on x > 1, and f (x) can be continuous on x = 1
f(x)1,lnx-x+11
Lnx1
Let's take n & # 178; (> 1) to replace the above formula X
lnn²