4
$\begingroup$

let $2\le a\le 13,a\in R$,and $x\in R$,show that: $$|1+x|^a\ge 1+ax+\dfrac{1}{1000}|x|^a\tag{1}$$

My try: let $$f(x)=|1+x|^a-1-ax-\dfrac{1}{1000}|x|^a$$

and since if $x>-1$,then $$|1+x|^a=(1+x)^a=1+ax+\dfrac{a(a-1)}{2}x^2+\cdots+x^a$$ and I fell this is nice reslut.because it is well konw this follow Bernoulli inequality

$$(1+x)^a\ge 1+ax,x>-1,a>1$$ But my inequality is strong than this .and I use computer test found this inequality $(1)$ is true.and I can't prove it.

BY the way I found in china book have this

enter image description here Thank you for you help

$\endgroup$
2
+25
$\begingroup$

Here is a partial answer : I show below that inequality (1) holds when $x\geq 0$ or $x \leq c_1=-\frac{1}{1-\big(\frac{1}{1000}\big)^{\frac{1}{13}}}$ (note that $c_1 \approx -2.42 \ldots$).

Let $g(x)=(1+x)^a-1-ax-x^a$ for $x\geq 0$. Then $g'(x)=a(1+x)^{a-1}-a-ax^{a-1}$, $g''(x)=a(a-1)\big((1+x)^{a-2}-x^{a-2}\big)$, so $g'$ is increasing, and hence $g'(x) \geq g'(0)=a(a-1) >0$, so $g$ is increasing, and hence $g(x) \geq g(0)=0$. So when $x\geq 0$ we have $|1+x|^a \geq 1+ax+|x|^a$, which is stronger than (1).

Next, if $x\leq c_1$ then $|x| \geq |c_1|$, $1-\frac{1}{|x|}\geq 1-\frac{1}{|c_1|}=\big(\frac{1}{1000}\big)^{\frac{1}{13}}$, so $\bigg(1-\frac{1}{|x|}\bigg)^a \geq \frac{1}{1000}$. It follows that $|1+x|^a \geq \frac{1}{1000}|x|^a$, which is also stronger than (1).

$\endgroup$
-1
$\begingroup$

The value of the function in 0 is 0 for every a. The function is continuous, so you could do the derivative and use binomial theorem to prove that for every x > 0 is positive and for every x < 0 is negative.

EDIT: http://i.imgur.com/CZcJutJ.jpg

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.