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.


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*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$

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.)
