Show that $x-\sqrt{x^2-x+1}<\frac{1}{2} $ for every real number $x$, without using differentiation Let $f$ be a function defined by :
$$f(x)=x-\sqrt{x^2-x+1}$$
Show that:
$$\forall x\in\mathbb{R},\quad f(x)<\dfrac{1}{2} $$
without use notion of différentiable 
let $x\in\mathbb{R}$
\begin{aligned}
f(x)-\dfrac{1}{2}&=x-\dfrac{1}{2}-\sqrt{x^2-x+1} \\
&=\dfrac{(x-\dfrac{1}{2})^2-|x^2-x+1|}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}\\
&= \dfrac{-3/4}{(x-\dfrac{1}{2})+\sqrt{x^2-x+1}}
\end{aligned}
i don't know if $(x-\dfrac{1}{2})+\sqrt{x^2-x+1} $ positive or negative  for all x in R

Notice that 
$$x-\sqrt{x^2-x+1}=x-\sqrt{(x-\frac{1}{2})^2+\frac{3}{4}}\\\sqrt{(x-\frac{1}{2})^2+\frac{3}{4}}> (x-\frac{1}{2})
$$
Now consider if $x<\frac{1}{2}$ the inequality is trivial,consider then $x\geq \frac{1}{2}$ then you're allowed to square the inequality and since $\frac{3}{4}>0$ the inequality is always true.Hence
$$ x-\sqrt{x^2-x+1}< \frac{1}{2}$$
