Finding the lowest value y can have in $y = \frac{1}{2}(e^x - e^{-x}) + \frac{n}{2}(e^x + e^{-x }) $ How can I find the lowest value $y$ can have when $n$ is greater than or equal to $2$ using only algebra?
$$y = \frac{1}{2}(e^x - e^{-x}) + \frac{n}{2}(e^x + e^{-x })$$
 A: Using AM-GM inequality
$$
y=\frac{(n+1)e^x+(n-1)e^{-x}}{2}\ge\sqrt{(n+1)(n-1)}=\sqrt{n^2-1}.
$$
Equality when $(n+1)e^x=(n-1)e^{-x}$ $\Rightarrow$ $x=\frac{1}{2}\ln\frac{n-1}{n+1}$.
A: Hint
The minimum or maximum would be reached when $\frac{dy}{dx}=0$. 
So, consider $$y =  \frac{1}{2}(e^x - e^{-x})  + \frac{n}{2}(e^x + e^{-x
})$$ Compute $\frac{dy}{dx}$, set it equal to $0$, solve for $x$ (which will be a function of $n$) and plug the value back in $y$.
Do not forget to use the second derivative test to verify that $y$ is at a minimum.
A: Solution without differentiation.
Given is
$$
y = \frac{1}{2} \Big( \exp(x) - \exp(-x) \Big)
+ \frac{n}{2} \Big( \exp(x) + \exp(-x) \Big).
$$

Step 1:
$$
y = \frac{n+1}{2} \exp(x) + \frac{n-1}{2} \exp(-x).
$$


Step 2:
$$
y = \frac{n+1}{2} \exp(-a) \exp(x+a) + \frac{n-1}{2} \exp(a) \exp(-[x+a]).
$$


Step 3:
Find $a$ such that
$$
\frac{n+1}{2} \exp(-a) : \frac{n-1}{2} \exp(a) = 1,
$$

so solve
$$
\frac{n+1}{n-1} \exp(-2a) = 1 \Rightarrow a = \log \sqrt{ \frac{n+1}{n-1} }.
$$

Step 4:
$$
y = \frac{\sqrt{n^2-1}}{2}
\Big[ \exp(x+a) + \exp(-[x+a]) \Big]
$$


Step 5:
$$
y = \sqrt{n^2-1}
\Bigg[ 1 + \Big[ \tfrac{1}{2} \exp(\tfrac{1}{2}[x+a])
- \tfrac{1}{2}\exp(-\tfrac{1}{2} [x+a]) \Big]^2  \Bigg]
$$

You can read out the minimum, as $y$ has the form $y = a ( 1 + \xi^2)$.
The minimum is $a$ for $\xi=0$.
The minimum is given by
$$
\sqrt{n^2-1}
$$
for
$$
x = \frac{1}{2} \log \frac{n-1}{n+1}.
$$
