Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between $a$ and $b$ 
Question:
Find the value of $n$ so that $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$ may be the geometric mean between $a$ and $b$.

My approach:
We know geometric mean between any two numbers $a$ and $b$ is given by $\sqrt{ab}$. By some processing in my brain I tried $n=-\frac{1}{2}$ and the result matched with the formula to find the geometric mean.
I wish to know is there any other disciplined approach for this problem?
Kindly guide me in this regard.
 A: Notice the following:
\begin{align*}
\forall a,b\in\mathbb{R}^+\,\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\sqrt{ab} &\iff \forall a,b\in\mathbb{R}^+a^{n+1}+b^{n+1}=a^n\sqrt{ab}+b^n\sqrt{ab}\\
&\iff \forall a,b\in\mathbb{R}^+ a^n\sqrt{a}(\sqrt{a}-\sqrt{b})+b^n\sqrt{b}(\sqrt{b}-\sqrt{a})=0\\
&\iff \forall a,b\in\mathbb{R}^+a^n\sqrt{a}(\sqrt{a}-\sqrt{b})=b^n\sqrt{b}(\sqrt{a}-\sqrt{b})
\end{align*}
A: Suppose $a>b$ 
$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab}\\ \implies a^{n+1} + b^{n+1}= a^{n+1/2}b^{1/2} + a^{1/2}b^{n+1/2} 
\\\implies a^{n+1/2}(a^{1/2}-b^{1/2}) - b^{n+1/2}(a^{1/2}-b^{1/2})=0\\
\implies (a^{1/2}-b^{1/2})(a^{n+1/2}-b^{n+1/2})=0
$
Since $a\ne b$, above equation forces $a^{n+1/2} = b^{n+1/2} \implies n+1/2=0$
A: Let $a=2b$ in  $$\frac{a^{n+1}+b^{n+1}}{a^n+b^n} = \sqrt{ab}$$
and simplify you get $$\frac {2^{n+1 }+1}{2^{n}+1} = \sqrt 2$$
$$ 2^{n+1}+1=2^{1/2}+2^{n+\frac {1}{2}} $$
The only solution is $n=-1/2$
A: Hint
For $a,b>0$ and $a\ne b$, the function$${a^{n+1}+b^{n+1}\over a^{n}+b^{n}}={a+b u^n\over 1+u^n}$$is absolutely increasing where $u={b\over a}$. There the answer $n=-{1\over 2}$ is the only such solution.
A: $\left(\frac{a^{n+1}+b^{n+1}}{a^n+b^n}\right)$ is G.M. between $a$ and $b$.
$\Leftrightarrow \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}=a^{(1 / 2)} b^{(1 / 2)} \quad\left[G . M\right.$. between $\left.a \& b=a^{(1 / 2)} b^{(1 / 2)}\right]$
$\Leftrightarrow a^{n+1}+b^{n+1}=a^{n+(1 / 2)} b^{(1 / 2)}+a^{(1 / 2)} b^{n+(1 / 2)}$
$\Leftrightarrow a^{n+1}-a^{n+(1 / 2)} b^{(1 / 2)}=a^{(1 / 2)} b^{n+(1 / 2)}-b^{n+1}$
$\Leftrightarrow a^{n+(1 / 2)}\left[a^{(1 / 2)}-b^{(1 / 2)}\right]=b^{n+(1 / 2)}\left[a^{(1 / 2)}-b^{(1 / 2)}\right]$
$\Leftrightarrow a^{n+(1 / 2)}=b^{n+(1 / 2)} \quad\left[\because a^{(1 / 2)}-b^{(1 / 2)} \neq 0\right.$, since $\left.a \neq b\right]$
$\Leftrightarrow\left(\frac{a}{b}\right)^{n+(1 / 2)}=1=\left(\frac{a}{b}\right)^{0}$
$\Leftrightarrow n+\frac{1}{2}=0$
$\Leftrightarrow n=-\frac{1}{2}$
