# 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

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

Answered by hamam_Abdallah on November 14, 2021