I need to prove the following:
$2\Phi(\frac{x-\mu}{\sigma})^2-3\Phi(\frac{x-\mu}{\sigma})+1 \leq 0$ for all $x\geq\mu$ where $\Phi(.)$ is the c.d.f. of a standard normal variable. I have done the following procedure:
The function attains the value of $0$ when $x=\mu$ and $x=\infty$ since:
$2\Phi(0)^2-3\Phi(0)+1=2(\frac{1}{2})^2-3(\frac{1}{2})=0$ and $\lim_{x \rightarrow \infty} 2\Phi(\frac{x-\mu}{\sigma})^2-3\Phi(\frac{x-\mu}{\sigma})+1=2-3+1=0$
Now I want to show that there is only one point in which the derivative of the function is zero (critical point) and that in this point the value of the function is negative. This implies that the function will never be positive in this interval.
$\frac{d}{dx} \Big(2\Phi(\frac{x-\mu}{\sigma})^2-3\Phi(\frac{x-\mu}{\sigma})+1\Big)=0$ implies $\Phi(\frac{x^*-\mu}{\sigma})=\frac{3}{4}$ which has a unique solution. For this $x^*$ the value of the function is $2(\frac{3}{4})^2-3(\frac{3}{4})+1=-\frac{1}{8}$.
Is my logic correct?
Thanks in advance
There is no need to go after derivatives, critical points etc.
Inequality $2y^2-3y+1=(2y-1)(y-1)\geq0$ can be rewritten as $\frac12\leq y\leq1$.
Applying that on $y=\Phi(\frac{x-\mu}{\sigma})$ we find $\frac12\leq \Phi(\frac{x-\mu}{\sigma})\leq1$.
This is equivalent with $x\geq\mu$.
This because $\Phi$ is a CDF that takes value $\frac12$ at $0$ and takes lower values on negative arguments.
On arguments in $[0,\infty)$ the CDF $\Phi$ will automatically take values in $[\Phi(0),1]=[\frac12,1]$.