Let $a$ and $b$ positive real numbers. Prove that
$$1 \leq \dfrac{\sqrt[n]{a^n+b^n}}{\sqrt[n+1]{a^{n+1}+b^{n+1}}} \leq \sqrt[n(n+1)]{2}.$$
Proposed by Ivan Borsenco.
Solution:
Since we can collect $a$ or $b$ in the given expression, it is sufficient to prove that
$$1 \leq \dfrac{\sqrt[n]{1+x^n}}{\sqrt[n+1]{1+x^{n+1}}} \leq \sqrt[n(n+1)]{2} \qquad \forall x \in \mathbb{R}^+,$$ or equivalently
$$1 \leq \dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n} \leq 2, \qquad \forall x \in \mathbb{R}^+.$$ Put $f(x)=\dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n}$. Then, $$f'(x)=\dfrac{n(n+1)x^{n-1}(1+x)^n(1+x^{n+1})^{n-1}(1-x)}{[(1+x^{n+1})^n]^2},$$ and $f'(x) > 0$ if and only if $x<1$, i.e. $f(1)$ is a maximum. Moreover, $f(0)=1$ and $\lim_{x \to +\infty} f(x)=1^+$, so $f(0) \leq f(x) \leq f(1)$ for all positive real numbers $x$, which gives $1 \leq f(x) \leq 2$ for all positive real numbers $x$.
$$1 \leq \dfrac{\sqrt[n]{1+x^n}}{\sqrt[n+1]{1+x^{n+1}}} \leq \sqrt[n(n+1)]{2} \qquad \forall x \in \mathbb{R}^+,$$ or equivalently
$$1 \leq \dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n} \leq 2, \qquad \forall x \in \mathbb{R}^+.$$ Put $f(x)=\dfrac{(1+x^n)^{n+1}}{(1+x^{n+1})^n}$. Then, $$f'(x)=\dfrac{n(n+1)x^{n-1}(1+x)^n(1+x^{n+1})^{n-1}(1-x)}{[(1+x^{n+1})^n]^2},$$ and $f'(x) > 0$ if and only if $x<1$, i.e. $f(1)$ is a maximum. Moreover, $f(0)=1$ and $\lim_{x \to +\infty} f(x)=1^+$, so $f(0) \leq f(x) \leq f(1)$ for all positive real numbers $x$, which gives $1 \leq f(x) \leq 2$ for all positive real numbers $x$.
No comments:
Post a Comment