How prove this two function which is bigger? let function 

$$f_{n}(x)=\left(1+x+\dfrac{1}{2!}x^2+\cdots+\dfrac{1}{n!}x^n\right)\left(\dfrac{x^2}{x+2}-e^{-x}+1\right)e^{-x},x\ge 0,n\in N^{+}$$

if $\lambda_{1},\lambda_{2},\mu_{1},\mu_{2}$ is postive numbers,and such
$\mu_{1}+\mu_{2}=1$
Question: following which is bigger:
$$f_{n}\left[(\lambda_{1}\mu_{1}+\lambda_{2}\mu_{2})\left(\dfrac{\mu_{1}}{\lambda_{1}}+\dfrac{\mu_{2}}{\lambda_{2}}\right)\right],  f_{n}\left[\dfrac{(\lambda_{1}+\lambda_{2})^2}{4\lambda_{1}\lambda_{2}}\right]$$
My try:
since
$$e^x=1+x+\dfrac{1}{2!}x^2+\cdots+\dfrac{1}{n!}x^n+\cdots+$$
But
this problem is 

$$1+x+\dfrac{1}{2!}x^2+\cdots+\dfrac{1}{n!}x^n$$
  so How prove it?Thank you 

This problem is from http://tieba.baidu.com/p/2682214392

If you are willing to rely on the problem setter to guarantee that there is a solution, you can do the following:  
If $\mu_1=\mu_2=\frac 12$ they are equal.
If $\mu_1=1, \mu_2=0$, the first is $f_n(1)$ and the second is $f_n(\frac 14(\frac {\lambda_1}{\lambda_2}+ \frac {\lambda_2}{\lambda_1}+2))$, which is $f_n$ of something greater than $1$.  As $f_n(x) \to \infty$ as $x \to \infty$, the second is larger.
We have not shown that this is true for all $n, \lambda$'s and $\mu$'s, but if there is a single answer, it must be this.
