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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

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    Corerect the numerator. Should be $ (x-1/2)^2$ – 2017-01-09

3 Answers 3

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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 $x \in \mathbb{R}$, $f(x) < \dfrac{1}{2}$.

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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 $$ x-\sqrt{x^2-x+1}< \frac{1}{2}$$

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    for second line you use : $$\forall x\in\mathbb{R},\forall a\in\mathbb{R}^+ \sqrt{x^2+a}>x $$ am i right ? – 2017-01-09
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    @Yacob should be $>|x|$, For kingW3, this should be $>|x-\frac12|$. – 2017-01-09
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    @zwim thank you so i can use it without proof it ? – 2017-01-09
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    @Yacob, the proof is esay: take $\sqrt.$ in the inequality $x^2+a>x^2$. – 2017-01-09
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    @Martín-BlasPérezPinilla i see thank you – 2017-01-09
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    Also the $|x|$ is not needed because if $x<0$ the inequality is trivially true – 2017-01-09
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    you're right, this is obvious now you say it, although I think this remark should go in the answer :p. – 2017-01-09
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    @zwim If $a>0$, then $\sqrt{x^2+a}>|x|$ and $|x|\ge x$. – 2017-01-09
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For $x\le 1/2$ is trivial. For $x > 1/2$: $$x - 1/2 < \sqrt{x^2-x+1}\iff (x-1/2)^2< x^2-x+1\iff 1/4 < 1.$$