IB HL Math, proving that a function is greater than $1$, for all $x>0$

Question

The function $f$ is given by $$f(x)= \frac {3^x + 1}{3^x - 3^{-x}}$$ for $x>0$. Show that $f(x)>1$ for all $x>0$

Hi all, I think I have solved this question but was having trouble proving this in a succinct and intuitive way. I would appreciate any help you might have to offer.

Thanks, T

Answer 1

With $t=3^x$, this is equivalent to $$\frac{t+1}{t-1/t}>1 $$ for all $t>1$. Some simple manipulations of the fraction expression take us to $$ 1<\frac{t+1}{t-1/t}=\frac{t(t+1)}{t^2-1}=\frac{t(t+1)}{(t+1)(t-1)}=\frac{t}{t-1}=1+\frac1{t-1}$$ an as $\frac1{t-1}$ is positive for $t>1$, we are done.

Answer 2

Well for $x>0$ we have $\begin{cases}3^x+1>3^x\\3^x-3^{-x}<3^x\end{cases}\ $ thus $\text{num}>\text{den}>0$ and the quotient is $>1$.

You can also do $(\text{num}-\text{den})=3^x+1-3^x+3^{-x}=1+3^{-x}>0$ for the same conclusion.