# prove that the inequality is positive for x-1>0 lne^2*(x^(x+1)) -2x]/(x+1)>=0

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We'll use the product rule for the first term of numerator:

lne^2*(x^(x+1)) = ln e^2 + ln x^(x+1)

We'll use the power property of logarithms:

lne^2*(x^(x+1)) = 2*ln e + (x+1)*ln x

But ln e = 1

lne^2*(x^(x+1)) = 2 + (x+1)*ln x

The inequality will become:

[2 + (x+1)*ln x - 2x]/(x+1) >= 0

We'll factorize by 2 the 1st and the last terms:

2(1-x)/(x+1) + (x+1)*ln x/(x+1) >= 0

ln x - 2(x-1)/(x+1) >= 0

In other words, we'll have to prove that the function f(x) = ln x - 2(x-1)/(x+1) is positive, for any value of x, higher than 1.

We'll evauate f(1) = ln 1 - 2(1-1)/(1+1) = 0 - 0 = 0

We'll re-write the inequality:

ln x - 2(x-1)/(x+1) >= f(1)

That means that f(1) is the minimum point of the function f(x).

**Therefore, for any value of x, locate din the set (1,+infinite), f(x)= [lne^2*(x^(x+1)) -2x]/(x+1) >=0.**