How to show that: if $n\ln\left(1+a/n\right)\geqslant k\ln\left(1+a/k\right)$ then $n\geqslant k$?

Question

Let $a>0$ and $n,k$ positive integers.

If $$n\ln\left(1+a/n\right)\geqslant k\ln\left(1+a/k\right),$$ then $$n\geqslant k.$$

I tried by contrapositive by I do not get much. If $n<k$ then I would have $$\ln\left(1+a/n\right)>\ln\left(1+a/k\right),$$ which does not help me a lot.

Have you tried bringing the leading terms in as a power and then showing $(1+a/n)^n \geq (1+a/k)^k$? — siegehalver, Mar 07 '16 at 20:14
It might be easier to go the other direction, assume $n \geq k$ and iff your way to the initial assumption. — siegehalver, Mar 07 '16 at 20:30

score 4 · Accepted Answer · answered Mar 07 '16 at 20:43

4

if $n\gt m\gt0$ and $a\gt0$, then Bernoulli's Inequality says $$ \left(1+\frac an\right)^{n/m}\gt\left(1+\frac an\frac nm\right)=\left(1+\frac am\right) $$ Therefore, $$ \left(1+\frac an\right)^n\gt\left(1+\frac am\right)^m $$ Reversing roles we have that for $m\gt n\gt0$, $$ \left(1+\frac an\right)^n\lt\left(1+\frac am\right)^m $$ Taking logarithms proves the required implication.

answered Mar 07 '16 at 20:43

robjohn

345,667

Good old Bernoulli's inequality - the swiss army knife of inequalities. – marty cohen Mar 07 '16 at 21:11
@martycohen: indeed! It is quite useful in cases like this. – robjohn Mar 07 '16 at 21:18
Though the version of BI proved by induction is for integer exponents. I'm not sure if the version you use can be proven as simply. I have supplied an answer here that uses calculus and works for any n and k, not just integers. – marty cohen Mar 07 '16 at 21:35
1

@martycohen: please take a look at this answer. It extends the integer version to rational exponents via induction. That is all we need here. Real exponents require continuity, but that follows from the definition of $a^x$ for real $x$. – robjohn Mar 07 '16 at 21:48
would the downvoter care to comment? – robjohn Mar 07 '16 at 21:52
very nice. upvoted. – marty cohen Mar 07 '16 at 22:02

score 3 · Answer 2 · answered Mar 07 '16 at 21:31

Let $f(x) =x \ln(1+a/x) $.

$\begin{array}\\ f'(x) &=\ln(1+a/x)+x(\ln(1+a/x))'\\ &=\ln(1+a/x)+x\frac{(1+a/x)'}{1+a/x}\\ &=\ln((x+a)/x)+x\frac{-a/x^2}{1+a/x}\\ &=-\ln(x/(x+a))-\frac{a}{x+a}\\ &=-\ln(1-a/(x+a))-\frac{a}{x+a}\\ &\gt a/(x+a)-\frac{a}{x+a} \qquad\text{since }-\ln(1-z) > z \text{ for }z > 0\\ &= 0\\ \end{array} $

Therefore $f(x)$ is increasing, which is what you want.

How to show that: if $n\ln\left(1+a/n\right)\geqslant k\ln\left(1+a/k\right)$ then $n\geqslant k$?

2 Answers2