Let [MATH]f(x)=\sqrt{1+x^2}-x.[/MATH] One can easily see that [MATH]f(x)\neq 0[/MATH], so either [MATH]f(x)>0[/MATH] or [MATH]f(x)<0[/MATH]. Since [MATH]\lim_{x\to -\infty}f(x)=\infty[/MATH] and [MATH]\lim_{x\to \infty}f(x)=0[/MATH], [MATH]f(x)\in(-\infty,0)[/MATH], whence [MATH]f(x)>0.[/MATH]