Let `x ,\ y` be two variables and `x >0,\ \ x y=1` , then minimum value of `x+y` is (a) 1 (b) 2 (c) `2 1/2` (d) `3 1/3`

Explanation

Let `f(x)=x+y`, where `x y=1`.


`\Rightarrow \quad f(x)=x+\frac{1}{x} `



`\Rightarrow \quad f^{\prime}(x)=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}}`


Also, `\quad f^{\prime \prime}(x)=\frac{2}{x^{3}}`
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