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

Question

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

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

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

Is this proof correct?

Answer 1

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

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,

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

Answer 2

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!

Answer 3

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 $0\leq y\leq x\leq1$ then: $$b^{\tfrac{m}{2}}-a^{\tfrac{m}{2}}=(x-y)\sum_{i=1}^{m}x^{m-i}y^{i-1}\leq(x-y)\frac{mx+my}{2}$$