Respuesta :

LammettHash

[tex]c+\dfrac1c=\dfrac{c^2+1}c=\dfrac{c^2-2c+1}c+2=\dfrac{(c-1)^2}c+2[/tex]

We have [tex](c-1)^2=0[/tex] when [tex]c=1[/tex], and for all other values of [tex]c[/tex] we have [tex](c-1)^2>0[/tex]. So

[tex]c+\dfrac1c=\dfrac{(c-1)^2}c+2\ge2[/tex]