Proving from first principles that if $0\lt x\lt x'\lt x+1$ then $2x(x'-x)\lt x'^2-x^2\lt 2(x+1)(x'-x)$

Question

Stuck on this analysis question...

Prove from first principles that for all real numbers $x,x'$ with $0\lt x\lt x'\lt x+1$:

$$2x(x'-x)\lt x'^2-x^2\lt 2(x+1)(x'-x)$$

I've solved the first order diff equation on the right hand side to get x=-1 and x=c1 e^x but I'm not sure if this was helpful or where to go from there...

Answer 1

Let $f(x)=x^2$. Then

$(x')^2-x^2=f(x')-f(x)=f'(z)(x'-x)$ with some $z \in (x,x').$

(Mean value theorem).

Then we have $2x <2z < 2x' <2(x+1)$, hence $2x <f'(z)<2(x+1).$

Answer 2

Rewrite the left inequality as: $(2x - (x'+x))(x'-x) = -(x-x')^2 <0$ which is clear since $x \ne x'$, and the right one: $(x'-x)(x'+x - 2x - 2)=(x'-x)(x'-x-2)< 0$ which is clear since $x' - x > 0, x'-x <1 < 2 \implies x'-x - 2 < 0$.

Answer 3

$$2x(x'-x)\lt x'^2-x^2\lt 2(x+1)(x'-x)$$ Since $x'-x>0$ $$2x\lt x'+x\lt 2(x+1)$$

$$x\lt x'\lt x+2$$

Which is true since it's true for $x+1$ on the right

Answer 4

You even have a tighter result: $$2x(x'-x)<x'^2-x^2<(2x+1)(x'-x).$$ Indeed rewrite $x'^2-x^2$ as $$(x'-x)(x'+x)=(x'-x)\bigl(2x+(x'-x)\bigr),$$ and use the hypothesis that $0<x'-x<1$.