# An Inequality with Fractions

### Solution 1

Rewrite $\displaystyle m\le\frac{a_k}{b_k}\le M\,$ as $mb_k\le a_k\le Mb_k\,$ and add all up:

$\displaystyle m\sum_{k=1}^nb_k\le M\sum_{k=1}^na_k\le\sum_{k=1}^nb_k,$

which is equivalent to

$\displaystyle m\le\frac{\displaystyle\sum_{k=1}^na_k}{\displaystyle\sum_{k=1}^nb_k}\le M.$

### Solution 2

The following statement may be considered as a reformulation of the above. It involves the notion of weighted averages:

Assume $x_k,\,$ $k\in\overline{1,n},\,$ are non-negative, $\omega_k,\,$ $k\in\overline{1,n},\,$ positive numbers such that $\displaystyle\sum_{k=1}^n\omega_k=1.\,$ Then

$\displaystyle\min_{k}x_k\le\sum_{k=1}^n\omega_kx_k\le\max_{k}x_k.$

The proof is practically the same as in Solution 1. Now, taking $\displaystyle x_k=\frac{a_k}{b_k}\,$ and $\displaystyle\omega_k=\frac{b_k}{\displaystyle\sum_{k=1}^nb_k}\,$ we obtain the latter.

### Solution 3

We can prove directly or refer to a property of the mediant fractions,viz.,

If $\displaystyle\frac{a_k}{b_k},\frac{a_{k'}}{b_{k'}}\in [m,M]\,$ then $\displaystyle\frac{a_k+a_{k'}}{b_k+b_{k'}}\in [m,M].$

Since $mb_k\le Mb_k\,$ and $mb_{k'}\le Mb_{k'}\,$ also $m(b_k+b_{k'})\le a_k+a_{k'}\le M(b_k+b_{k'}),\,$ which proves the above and the rest flows by induction.

### Application 1

Assume $\displaystyle 0\lt\alpha_1\le\alpha_2\le\ldots\le\alpha_n\le\frac{\pi}{2}.\,$ Then

$\displaystyle \tan\alpha_1\le\frac{\sin\alpha_1+\sin\alpha_2+\ldots+\sin\alpha_n}{\cos\alpha_1+\cos\alpha_2+\ldots+\cos\alpha_n}\le\tan\alpha_n.$

Indeed, since $y=\tan x\,$ is strictly increasing on $\displaystyle\left(0,\frac{\pi}{2}\right),\,$ then, for every $k\in\overline{1,n},\,$ $\displaystyle\tan\alpha_1\le\tan\alpha_k=\frac{\sin\alpha_k}{\cos\alpha_k}\le\tan\alpha_n.\,$ The above inequality applies directly.

### Application 2

Assume $\displaystyle 0\lt\alpha_1,\alpha_2,\ldots,\alpha_n\le\frac{\pi}{6}.\,$ Then

$\displaystyle \frac{3}{\pi}\le\frac{\sin\alpha_1+\sin\alpha_2+\ldots+\sin\alpha_n}{\alpha_1+\alpha_2+\ldots+\alpha_n}\le 1.$

Function $\displaystyle y=\frac{\sin (x)}{x}\,$ is strictly decreasing on $\displaystyle\left( 0,\frac{\pi}{2}\right)\,$ with $f(0)=1,\,$ such that, for any $k\in\overline{1,n},\,$ $\displaystyle\frac{3}{\pi}=\frac{\displaystyle\sin\frac{\pi}{6}}{\displaystyle\frac{\pi}{6}}\le\frac{\sin\alpha_k}{\alpha_k}\le 1\,$ and the general inequality applies.

### Application 3

Let $\displaystyle m\le\sqrt[p_k]{a_k}\le M.\,$ Then

$\displaystyle m\le\sqrt[\displaystyle\sum_{k=1}^np_k]{\prod_{k=1}^na_k}\le M.$

Indeed, it is given that $\displaystyle \ln (m)\le\frac{\ln (a_k)}{p_k} \le \ln (M).\,$ It follows that

$\displaystyle \ln (m)\le\frac{\displaystyle\sum_{k=1}^n\ln (a_k)}{\displaystyle\sum_{k=1}^np_k}=\frac{\displaystyle \ln \left(\prod_{k=1}^na_k\right)}{\displaystyle\sum_{k=1}^np_k}=\ln\left(\left(\displaystyle\prod_{k=1}^na_k\right)^{\displaystyle\frac{1}{\displaystyle\sum_{k=1}^np_k}}\right)\le \ln (M).$

### Application 4

For $x\ge 0,$

$\displaystyle\frac{1}{n}\le\frac{1+2x+\ldots+nx^{n-1}}{n+(n-1)x+\ldots+x^{n-1}}\le n.$

### Application 5

For $x\ge 0,$

$\displaystyle\frac{1}{n}\le\frac{1+2x+\ldots+nx^{n-1}}{1+2^2x+\ldots+n^2x^{n-1}}\le 1.$

### Acknowledgment

The problem, Applications 1 and 3 came from an old Russian collection of problems by V. A. Krechmar. Solution 2 has been suggested by Faryad D. Sahneh; Solution 3 is by N. N. Taleb. Dorin Marghidanu has passed on a page (204) from D.S. Mitrinovic's book "Analyttic inequalities," (Springer, 1970) with a statement and a proof of the inequality. Applications 4 and 5 come from that book.