# 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.

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.

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.

• Some comments : a factor $4$ is missing in your definition of $B$, there’s no need to restrict $f$ to $[1,\infty)$, $f$ is already decreasing on $[0,\infty)$ and one can show it without any approximation at all (see my answer which picks up where you stopped). Nov 6, 2013 at 7:12