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
 A: 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.
A: 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. 
A: Let $A = (\lambda_1 \mu_1 + \lambda_2 \mu_2)(\mu_1/\lambda_1 + \mu_2/\lambda_2)$ and $B= (\lambda_1 + \lambda_2)^2/ \lambda_1 \lambda_2$. Then you can compute explicitly that:
$$ 
A-B = \mu_1^2 + \mu_2^2 + \mu_1 \mu_2 (\lambda_1^2 + \lambda_2^2)/\lambda_1\lambda_2 - (\mu_1+\mu_2)^2 (\frac{1}{2} + (\lambda_1^2 + \lambda_2^2)/4\lambda_1\lambda_2 ) \\
= \frac{(\mu_1-\mu_2)^2}{2} -  (\mu_1-\mu_2)^2(\lambda_1^2 + \lambda_2^2)/4\lambda_1\lambda_2 \\
= - (\mu_1-\mu_2)^2(\lambda_1 - \lambda_2)^2/4\lambda_1\lambda_2 < 0
$$
Hence, $B \geq A$ in general. It is fairly easy to show that $A\geq 1$, and  I am reasonably sure you can actually find $\mu$'s and $\lambda$'s such that $A,B$ take any prescribed values satisfying these constraints. Thus, what we need to show is basically that $f$ is monotoneous on $[1,\infty)$. 
For this, you can use brute force. Differentiate $f_n$ and see if the result is positive. Unless I did a computational error (which is probably the case), the derivative comes out as:
$$ \frac{x e^{-x}}{(x+2)^2} \left( (1+...x^n/n!) - \frac{1}{n!} (x(x+2) - \text{sth small}) \right)$$
This can (hopefully) be seen to be positive by some rough approximations.
