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

As shown in Feanor’s answer, if $A=(\lambda_{1}\mu_{1}+\lambda_{2}\mu_{2})
\left(\dfrac{\mu_{1}}{\lambda_{1}}+\dfrac{\mu_{2}}{\lambda_{2}}\right)$ and
$B=\dfrac{(\lambda_{1}+\lambda_{2})^2}{4\lambda_{1}\lambda_{2}}$, then 
$A \leq B$ because of the algebraic identity
$$
(y_1+y_2)^2B-A=\frac{(\lambda_2-\lambda_1)^2(\mu_2-\mu_1)^2}{4\lambda_1\lambda_2}
$$
I show below that $f_n$ is decreasing (which entails $f_n(A) > f_n(B)$, answering 
your question).
For $x> 0$, let
$$
g_{n}(x)=\left(1+x+\dfrac{1}{2!}x^2+\cdots+\dfrac{1}{n!}x^n\right)e^{-x}, \ \ \
h_{n}(x)=\dfrac{x^2}{x+2}-e^{-x}+1
$$
We have $h'_{n}(x)=1-\frac{4}{(x+2)^2}+e^{-x}$ and both 
$1-\frac{4}{(x+2)^2}$ and $e^{-x}$ are positive. So $h_n$ is increasing.
Since $h_n(0)=0$, we see also that $h_n$ is positive.
We have $g'_{n}(x)=-\frac{x^n}{n!}e^{-x}$, so $g_n$ is decreasing (and $g_n$
is obviously positive).
To conclude, $f_n$ is the product of two positive functions, one of whom
is decreasing and the other is increasing, so $f_n$ is decreasing. 
