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.
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