Proving $|(x-x_0)(x-x_1)|\leq \frac {1}{4}(x_1-x_0)^2$

Question

I need to prove that $$ |(x-x_0)(x-x_1)|\leq \frac {1}{4}(x_1-x_0)^2 $$ for all $x\in < x_0 ; x_1 >$

The only thing I have notived it that the expression with absoltue value is always non-positive, so the inequality is equvalent to $$ -(x-x_0)(x-x_1)\leq \frac {1}{4}(x_1-x_0)^2 $$ but I'm stuck here

@dmtri so I have achieved form $4x(x_1-x+x_0)\leq (x_0+x_1)^2$ and that I can prove using $4xy\leq (x+y)^2$? Correct? — MartinYakuza, Apr 23 '20 at 17:18

score 2 · Answer 1 · answered Apr 23 '20 at 19:49

2

For $ x \in (x_0, x_1)$, you have $(x-x_0) + (x_1-x) = x_1-x_0$.

Then prove (it's not difficult) that the product of two numbers whose sum is constant is maximum when those two numbers are equal.

Apply that to your case, $\left\vert(x-x_0)(x-x_1) \right\vert$ is maximum when $x-x_0 = x_1-x = \frac{x_1-x_0}{2}$ and therefore the maximum is indeed equal to $\frac{(x_1-x_0)^2}{4}$.

answered Apr 23 '20 at 19:49

mathcounterexamples.net

70,018

proving that fact you mentioned is that when we have sum of two numbers equal s, then product is x(s-x) so we have roots in 0 and s, so the max value is halfway between them so it's in s/2, right? – MartinYakuza Apr 23 '20 at 20:02
Yes. You can also make a proof by taking the derivative of $g(x)=x(s-x)$. – mathcounterexamples.net Apr 23 '20 at 20:03
Yes, but I don't really see the point of derivating parabola to find it's maximum – MartinYakuza Apr 23 '20 at 20:09
$g(x)=(x-x_0)(x-x_1) $ is the equation of a parabola! – mathcounterexamples.net Apr 23 '20 at 20:14
I imagine that you know that to find the extrema of a function, its derivative is important?!? – mathcounterexamples.net Apr 23 '20 at 20:18
Yeah, but it's a well know fact, that the extrema of parabola is in it's vertex – MartinYakuza Apr 23 '20 at 20:20

score 1 · Accepted Answer · answered Apr 23 '20 at 20:59

1

Note

$$0\le (x-\frac{x_0+x_1}2)^2$$

and $$( \frac{x_1-x_0}2)^2 - (x-\frac{x_1+x_0}2)^2 \le ( \frac{x_1-x_0}2)^2 $$

Thus,

$$(x_1-x)(x-x_0) \le \frac14( x_1-x_0)^2 $$

answered Apr 23 '20 at 20:59

Quanto

97,352

could you also help me with question, that is generalisation of this problem? https://math.stackexchange.com/questions/3640493/proving-fracb-ab-x-cdot-fracb-ak1k1-geq-fracb-akk1 – MartinYakuza Apr 23 '20 at 21:02
@MartinYakuza - will take a look – Quanto Apr 23 '20 at 21:03

Proving $|(x-x_0)(x-x_1)|\leq \frac {1}{4}(x_1-x_0)^2$

2 Answers2