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Let $0leq a leq b leq 1$. Then we have for all natural numbers $mgeq 2$ the inequality $b^{frac m2}-a^{frac m2} leqfrac m2(b-a)$

Let $0leq a leq b leq 1$. Then we have for all natural numbers $mgeq 2$ the inequality $b^{frac{m}{2}}-a^{frac{m}{2}} leqfrac{m}{2}left(b-aright)$.

My first idea was to consider the function $f(x)=x^{frac{m}{2}-1}$ on the interval $[0,1]$. Since $mgeq 2$ it follows that $underset{xin [0,1]}{text{sup}}f(x)=1.$ Then, by the fundamental theorem of calculus we can conclude:

$begin{align*} b^{frac{m}{2}}-a^{frac{m}{2}} & =displaystyleint_{a}^{b}frac{m}{2}f(x),dx \ & =frac{m}{2} displaystyleint_{a}^{b} x^{frac{m}{2}-1},dx \
& leq frac{m}{2} underset{xin [0,1]}{text{sup}}f(x)(b-a) \
& = frac{m}{2}(b-a) end{align*}$

Is this proof correct?

Mathematics Asked by Giuliano Cantina on November 14, 2021

3 Answers

3 Answers

Other way is: if $x=sqrt{b}, y=sqrt{a}$, then: $$b^{tfrac{m}{2}}-a^{tfrac{m}{2}}=(x-y)sum_{i=1}^{m}x^{m-i}y^{i-1}$$ How $0leq yleq xleq1$ then: $$b^{tfrac{m}{2}}-a^{tfrac{m}{2}}=(x-y)sum_{i=1}^{m}x^{m-i}y^{i-1}leq(x-y)frac{mx+my}{2}$$

Answered by AsdrubalBeltran on November 14, 2021

There is also the following reasoning.

Let $f(b)=frac{m}{2}(b-a)-b^{frac{m}{2}}+a^{frac{m}{2}}.$

Thus, $$f'(b)=frac{m}{2}-frac{m}{2}cdot b^{frac{m}{2}-1}=frac{m}{2}left(1-b^{frac{m-2}{2}}right)geq0.$$ Id est, $$f(b)geq f(a)=0$$ and we are done!

Answered by Michael Rozenberg on November 14, 2021

Your proof is correct but you had to say that $$sup_{[a,b]}fle sup_{[0,1]}f$$

hint

here is an other :

Let $$g(x)=x^{frac m2}$$ $ g $ is continuous at $ [a,b] $ and differentiable at $ (a,b) $, then by MVT,

$$exists cin (a,b);; :;frac{g(b)-g(a)}{b-a}=g'(c)$$ but $$g'(c)=frac m2 c^{frac m2-1}$$ and $$0<c<1$$

Answered by hamam_Abdallah on November 14, 2021

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