Show that $(x^n+y^n)^{\frac{1}{n}}$ is monotone where $0\leq x<y$

Question

Let $x,y$ such that $0\leq x<y$. I'm having problem in at proving that the sequence $a_{n}=(x^{n}+y^{n})^{\frac{1}{n}}$ is monotone. I tried it using the function $f(z)=(x^z+y^z)^{\frac{1}{z}}$ and see that it has negative derivative but I couldn't. I also tried on using induction on $n$ to prove that $a_{n}$ is decreasing but neither did I can. (sorry for my English). Do you have any sugestion?

I already proved that this sequence is bounded. I also know how to prove the convergence using sandwich theorem and that this sequence converges to $\max\{x,y\}$ but what I want here is to prove only the monotony.

Answer 1

Let $n\in\mathbb{N}$. For $0< t<1$, $1+t^{n+1}<1+t^n$. The map $x\mapsto x^{1/n}$ is monotone increasing; hence $$(1+t^{n+1})^{1/n}<(1+t^n)^{1/n}$$ As $a:=1+t^{n+1}>1$, the map $x\mapsto a^x$ is monotone increasing; hence $$(1+t^{n+1})^{1/{(n+1)}}< (1+t^{n+1})^{1/n}<(1+t^n)^{1/n}$$

Answer 2

If $0\lt x\lt y$, then $$ \left(x^n+y^n\right)^{1/n}=y\left(1+\left(\frac xy\right)^n\right)^{1/n}\tag1 $$ Let $u=\frac xy$. Then $0\lt u\lt1$. Furthermore, $$ \begin{align} \left(\frac{\left(1+u^{n+1}\right)^{\frac1{n+1}}}{\left(1+u^n\right)^{\frac1n}}\right)^{n(n+1)} &=\frac{\left(1+u^{n+1}\right)^n}{\left(1+u^n\right)^{n+1}}\tag{2a}\\ &=\left(\frac{1+u^{n+1}}{1+u^n}\right)^n\frac1{1+u^n}\tag{2b}\\[11pt] &\lt1\tag{2c} \end{align} $$ Explanation:
$\text{(2a)}$: raise the ratio of consecutive terms to the $n(n+1)$ power
$\text{(2b)}$: pull out a factor of $\frac1{1+u^n}$ and collect the exponent of $n$ over the ratio
$\text{(2c)}$: $\frac{1+u^{n+1}}{1+u^n}\lt1$ and $\frac1{1+u^n}\lt1$

Therefore, $\left(1+u^n\right)^{1/n}$ is decreasing, and thus, $\left(x^n+y^n\right)^{1/n}$ is decreasing.

If $x=0$, then $\left(x^n+y^n\right)^{1/n}=y$ is constant in $n$.

Answer 3

Let $f(x)=x^{1+\frac{1}{n}}$, $x^n=a$ and $y^n=b$.

Thus, $f$ is a convex function and $(a+b,0)\succ(b,a).$

Id est, by Karamata $$f(a+b)+f(0)\geq f(b)+f(a)$$ or $$(x^n+y^n)^{1+\frac{1}{n}}\geq y^{n+1}+x^{n+1},$$ which says that our sequence decreases.

Answer 4

$a_n$ is monotone iff $\log a_n = \frac{1}{n}\log(x^n+y^n)$ is monotone. Let us set $\frac{y}{x}=e^z$. By the hyphotesis we have $z>0$ and $a_n = \log(x)+\frac{1}{n}\log(1+e^{nz})$, so it is enough to show that $\frac{1}{n}\log(1+e^{nz})$ is monotone with respect to $n$. Regarding $r$ as a variable in $\mathbb{R}^+$ we have

$$ r^2\cdot \frac{\partial}{\partial r}\left(\frac{1}{r}\log(1+e^{zr})\right) = \frac{rz e^{rz}}{1+e^{rz}}-\log(1+e^{rz})$$ so it is enough to show that $$ \frac{s e^s}{1+e^s} < \log(1+e^s) $$ holds for any $s>0$. This is equivalent to $$ \int_{-\infty}^{s} \frac{u e^u}{(e^u+1)^2}\,du < 0 $$ which is obvious since the integrand function is odd and it is positive over $\mathbb{R}^+$.