# proving question

Let x be a positive number, with $x^2>2.$
Let $y = x/2 + 1/x$
Show that $0, and that $y^2>2$. Explain why this shows that there is no smallest positive number whoes square is more than 2.
Just do the claims in turn. To show that $y < x$, show that $1/x < x/2$. To show $y^2>2$, calculate $(x/2+1/x)^2$. I hope you can figure out the last part provided the rest is shown.