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Discussion Response Post to: question based on deriving an ineqaulity

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Titun (1467)

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Hiiiiii,

Although this problem lies in the Algebra section, I would like to solve this problem using calculus.

Let us define a function, f(x) = [(x-c)/(x+c)]^x where x>c and x,c>0

It is obvious f(x)>0 for all x>c>0 If now, I can show f ' (x) >0 i.e f(x) is an increasing function, then it is obvious that

f(a) > f(b) where a>b>c>0 i.e [(a-c)/(a+c)]^a > [(b-c)/(b+c)]^b where a>b>c>0

ln f(x) = x[ln(x-c)-ln(x+c)]

Differentiating both sides w.r.t x, we have,

f ' (x) = [(x-c)/(x+c)]^x [ln{(x-c)/(x+c)} + 2cx/(x^2 - c^2)]

[(x-c)/(x+c)]^x >0 ; so our proof reduces to proving only that

[ln{(x-c)/(x+c)} + 2cx/(x^2 - c^2)] > 0

ln{(x-c)/(x+c)} = ln{(1-y)/(1+y)} where y=c/x, so 0<y<1

Now, ln(1+y) = y - y^2 / 2 + y^3 / 3 ..... (using logarithmic series formula)

Ln(1-y) = -y - y^2 / 2 - y^3 / 3 ......

Therefore, ln(1-y)-ln(1+y) = ln{(1-y)/(1+y)} = -2[y+y^3 / 3 + y^5 / 5 + ..... ]

i.e. ln{(1-y)/(1+y)} > - 2 [ y + y^3 + y^5 + ...... ]

i.e. ln{(1-y)/(1+y)} + 2y/(1-y^2) >0

Putting y = c/x, we have, [ln{(x-c)/(x+c)} + 2cx/(x^2 - c^2)] > 0

Hence, f ' (x) > 0 i.e f(x) is an increasing function. So, f(a) > f(b) for a>b>c>0

i.e. [(a-c)/(a+c)]^a > [(b-c)/(b+c)]^b

I hope I have explained all the steps properly. Also plz tell me if I went wrong anywhere.

You never know what is enough till you know what is more than enough. Titun

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