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
 A: 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
$$ x-\sqrt{x^2-x+1}< \frac{1}{2}$$
A: For $x\le 1/2$ is trivial. For $x > 1/2$:
$$x - 1/2 < \sqrt{x^2-x+1}\iff (x-1/2)^2< x^2-x+1\iff 1/4 < 1.$$
A: Note that this proof is a bit longer than expected, but it shows some line of thoughts....We have: If $x \le 0 \implies f(x) \le 0 < \dfrac{1}{2} \implies f(x) < \dfrac{1}{2}$. Thus if $x > 0 \implies f(x) = \dfrac{x-1}{x+\sqrt{x^2-x+1}}$. Here we can reason a little bit. If $0 < x \le 1 \implies f(x) \le 0 \implies f(x) < \dfrac{1}{2}$. Thus $x > 1$. If again $1 < x < 2\implies x^2 - x + 1 > x^2-4x+4 \implies x^2-x+1 > (x-2)^2\implies \sqrt{x^2-x+1} > |x-2| = 2-x\implies f(x) < \dfrac{x-1}{x+2-x}= \dfrac{x-1}{2} < \dfrac{1}{2}$ since $x < 2$. Finally, if $x \ge 2\implies |x-2| = x-2\implies f(x) < \dfrac{x-1}{x+x-2}= \dfrac{x-1}{2x-2}= \dfrac{x-1}{2(x-1)} = \dfrac{1}{2}$. Thus for all $x \in \mathbb{R}$, $f(x) < \dfrac{1}{2}$.
