PROBLEM 1. a) Determine the value of the number $x$, if

$$
\frac{2014}{2015 \cdot 2015-2015-2015+1}=\frac{x}{2014 \cdot 2015}
$$

b) Using the inequality $\frac{1}{a}+\frac{1}{b}>\frac{4}{a+b}$, for any positive and distinct $a$ and $b$, show that

$$
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{499}+\frac{1}{500}>\frac{13}{7}
$$
## PROBLEM 1.

(2p) a) $2015 \cdot 2015 - 2015 - 2015 + 1 = 2014 \cdot 2014$

(2p) \quad The solution is $x = 2015$

(1p)

b) We have $1 + \frac{1}{500} > \frac{4}{501}, \frac{1}{2} + \frac{1}{499} > \frac{4}{501}, \ldots, \frac{1}{250} + \frac{1}{251} > \frac{4}{501}$

(1p) from which, $1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{499} + \frac{1}{500} > 250 \cdot \frac{4}{501} = \frac{1000}{501}$

(1p) Since $\frac{1000}{501} > \frac{13}{7}$, we obtain the required inequality.