Need help in deducing an inequality related to natural logarithms

Question

I am trying an question in analysis whose solution depends upon deriving an inequality .

If $x < a < 1/2$ then prove that $\log (x+1) -2 \log(x+2) < \log(a+1) - 2 \log(a+2)$ .

I tried using maxima minima but i cannot derive it. I am getting opposite result.

Can someone please help.

Edit-> I already had an idea that there could be a mistake ,but I was thinking that it might be a possibility of me being wrong. I have found a way of doing it without this inequality. I request you all to ignore the question and thanks for giving your time.

Answer 1

Throw stones at it until something falls over.

$\log (x+1) -2 \log(x+2)=\log(x+1) - \log(x+2)^2 = \log \frac {x+1}{(x+2)^2}$

And $ \log(a+1) - 2 \log(a+2) = \log\frac {a+1}{(a+2)^2}$.

Now $e > 1$ and we know that $m < n \iff e^m < e^n$ (we do know that right? Take a moment to convince yourself of that). So that means $\log w < \log z \iff w < z$ (just let $e^m = w$ and $e^n = z$).

So the statement $\log (x+1) -2 \log(x+2) < \log(a+1) - 2 \log(a+2)$ is true if and only if

$\frac {x+1}{(x+2)^2} < \frac {a+1}{(a+2)^2}$

Okay..... how do we get that from $x < a < \frac 12$?

As $\log x+1$ and $\log a+1$ are defined we know that $x+1 > 0$ and $a+1 > 0$ so that is true if and only if

$\frac {x+1}{a+1} < (\frac {a+2}{x+2})^2$

And we have $-1< x < a < \frac 12$

$\frac {x+1}{a+1} = \frac {a+1}{a+1} - \frac {a-x}{a+1}=1-\frac {a-x}{a+1}<1$.

And $\frac {a+2}{x+2} = \frac {x+2}{x+2} + \frac {a-x}{x+2} = 1 +\frac {a-x}{a+2} > 1$

So $\frac {x+1}{a+1} < 1 < (\frac {a+2}{x+2})^2$.

And that's it. We are done.

(.... Hmm, unless I made an error. I never used $a < \frac 12$.... which tells you that $\log (x+1), \log(x+2), \log(a+1),$ and $\log(a+2)$ are all negative.. which I didn't need.... maybe..... When I said $m<n \iff e^m < e^n$ one may mistakenly assume $m> 0$ but it is still true if $n < 0$. $m < n < 0 \iff 0 < e^m < e^n < 1$.)

Answer 2

In order for all of the logarithm expressions to make sense, we must assume $a > x > -1.$

Write $p = a + 1$ and $q = x + 1.$ Then $p > q > 0,$ and: $$ \frac{a+1}{(a+2)^2} - \frac{x+1}{(x+2)^2} = \frac{p}{(p+1)^2} - \frac{q}{(q+1)^2} = \frac{pq^2 - qp^2 + p - q}{(p+1)^2(q+1)^2} = \frac{(p-q)(1-pq)}{(p+1)^2(q+1)^2}. $$ We know that $p - q > 0$ and $(p+1)^2(q+1)^2 > 0,$ and we want to have: $$ \frac{a+1}{(a+2)^2} > \frac{x+1}{(x+2)^2} $$ so that we can get the required result by taking logarithms of both sides.

Therefore, we want to have $1 - pq > 0,$ or equivalently $pq < 1,$ or again $(a + 1)(x + 1) < 1.$

Therefore, if the condition $x < a$ isn't a misprint, we must have $x < 0.$

At the very least, then, the problem is lacking some required condition.

Answer 3

Assume $-1<x$ (otherwise the log is not a real number) and $x<a<1/2$. The problem is equivalent to (since the log is a monotonically increasing function) $$ f(a) > f(x) $$ where $$ f(z) = \frac{z+1}{(z+2)^2}. $$ The function $f(z)$ has a maximum in $z=0$: it is increasing for $z<0$ and decreasing for $z>0$. Hence, when $a$ and $x$ are both negative, you have that $f(a)>f(x)$ if $a>x$ (because $f$ is increasing). This gives you a first solution: $f(a)>f(x)$ when $-1<x<a<0$.

Now consider the case $a>0$: it is easy to see that you must have $-1 < x < -a/(1 + a)$ to guarantee that $f(a)>f(x)$.

Summarizing: you have that $f(a)>f(x)$ when $$-1 < a \leq 0 \quad \& \quad -1 < x < a$$ or when $$ 0 < a < 1/2 \quad \& \quad -1 < x < -a/(1 + a)$$