transforming $\chi^2(n)$ distribution to $t$ distribution if we have two independent random variables for both of them $\chi^2(n)$ distribution 
Prove it has a t distribution with degree freedom ($n$)
$$\frac{\sqrt n (x_1 +x_2 )}{ \sqrt {x_1 x_2}}$$

@MichaelHardy and @Henry are trying to explain that you cannot
prove this because it is not true.
Here is something that is true. It is valuely related to what
you posted and might even be what you intended. It illustrates the definition of
a random variable with Student's t distribution:
Let $X_1 \sim Norm(0,1)$ and, independently, $X_2 \sim Chisq(df=n).$
Then $$T = \frac{\sqrt{n}X_1}{\sqrt{X_2}} = \frac{X_1}{\sqrt{X_2/n}} \sim T(n),$$
Student's t distribution with $n$ degrees of freedom. You can find the
proof in many mathematical statistics texts.
The simulation below in R statistical software demonstrates this relationship
for $df = 15.$
m = 10^6;  df = 15;  x1 = rnorm(m);  x2 = rchisq(m, df)
t = sqrt(df)*x1/sqrt(x2)
mean(t);  var(t);  df/(df-2)
## -0.0001963736    # aprx E(T) = 0
## 1.15594          # aprs Var(T) = 15/13
## 1.153846         # exact Var(T)
quantile(t)         # min, aprx quartiles, max
##            0%           25%           50%           75%          100% 
## -1.012026e+01 -6.929461e-01 -5.062215e-04  6.930493e-01  7.579955e+00 
qt(c(.25,.75), df)
## -0.6911969  0.6911969  # exact lower and upper quartiles of T(15)

Below is a histogram of the million realizations of $T$ simulated in
this way. Student's t distributions tend to have heavy tails, hence the
'long' horizontal axis to accommodate a few stragling values far from 0
in both directions (which produce bars too short to show at the resolution
of the graph). The curve is the PDF of $T(15).$

