Prove that $a+\frac{1}{b}>2$ or $b+\frac{1}{a}>2$ for two strict positive numbers Another Olympiad Problem, let $x$ and $a$ and $b$ be strictly real positive numbers.


*

*Prove that   $x$+$\frac{1}{x}$$>$$2$ (proven)

*Than conclude that   $a$+$\frac{1}{b}$$>$$2$  or 
    $b$+$\frac{1}{a}$$>$$2$


For the second question, I know I am supposed to replace $x$ with a number in function of $a$ and $b$ , but I can't find it .
if we replace $x$ by $ab$ we still have a problem.
$ab$+$\frac{1}{ab}$$+2>$$4$ , we can't just conclude that $a$+$\frac{1}{b}$$>$$2$  or         $b$+$\frac{1}{a}$$>$$2$
 A: Hint: Note that 
$$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right)=ab+\frac{1}{ab}+2.$$
A: Your first statement $x + \frac 1x > 2$ is not fully correct, you have
$$
 x + \frac 1x \ge 2
$$
for $x > 0$, with equality exactly for $x = 1$.
And as the example $a=b=1$ shows, you can only conclude that
$$ 
a+\frac{1}{b} \ge 2 \quad \text{ or } \quad b+\frac{1}{a} \ge 2 \, .
$$
which is your claim with $\ge$ instead of $>$.
This follows from (similar to André's solution):
$$
\left(a+\frac{1}{b}\right) + \left(b+\frac{1}{a}\right)= a + \frac 1a + b + \frac 1b \ge 2 + 2 = 4
$$
(with equality exactly for $a=b=1$),
so at least one term on the left-hand side must be greater or equal
to $2$.
If $a$ and $b$ are not both equal to one then you can conclude that
$$ 
a+\frac{1}{b} > 2 \quad \text{ or } \quad b+\frac{1}{a} > 2 \, .
$$
A: Slightly different.
If $a = b$ then $a + 1/b =  b + 1/a =a + 1/a \ge 2$.
If $a > b$ then $a + 1/b > a + 1/a \ge 2$.
If $a < b$ then $b + 1/a > b + 1/b \ge 2$.
(If $a = b = 1$ then $a + 1/b = 2$. Otherwise $a + 1/b > 2$ or $b + 1/a > 2$ or both.)
A: Since you already settled the case $a=b$ in the first part of the question, we may assume without loss of generality that $a<b$.     
Therefore we have $b+\dfrac{1}{a} > b +  \dfrac{1}{b} \geq 2$, which yields the required result.                                                                                 
