Any useful tips and tricks when trying to determine if one expression is greater than or equal to another expression? I tend to get stuck on questions that ask to determine if a given expression is greater than another expression. For example:
Let:
$MSE_\alpha = \frac{1}{5}\pi(1-\pi)$
$MSE_\beta = \frac{1}{36}5\pi(1-\pi) + (\frac{1}{12}-\frac{1}{6}\pi)^2$
where $0<\pi<1$ and $MSE$ stands for mean squared error.
How do I systematically determine which MSE is the minimum between the two?
The first thought that crossed my mind is to deduct one of the MSE from the other and see if the result is greater than or equal to zero but that seems like a long winded way of doing things (So far, all the answers I have seen did not approach the problem this way too).
So, I decided to look at each equation carefully and note the following:


*

*The second term in $MSE_\beta$ will always be more than zero since the term is squared

*$\frac{1}{5} > \frac{5}{36}$, so $MSE_\beta$ can be greater than $MSE_\alpha$ if the second term of $MSE_\beta$ is large enough.
Is that all I can conclude? Please tell me how you would go about solving this kind of problems and the answer to this specific problem too.
 A: For unknown $\pi$, the problem is to solve the sign of $MSE_\alpha(\pi) - MSE_\beta(\pi)$. Because both terms are 2nd degree polynomials, the problem reduces to finding how the sign of a 2nd degree polynomial behaves as a function of the variable ($\pi$ in this case). The OP asked for a "systematical" method so I will show the general case without plugging in any values.
Let $p(x)=ax^2+bx+c$ be a polynomial. Because $p$ is continous, its sign can change only at a zero of $p$ (but it does not have to change at a zero!).
The case $a=0$ is quite trivial so I'll omit it here. Now we note that the 2nd degree term dominates $p(x)$ for large $|x|$. Therefore, $p(x)$ has the same sign as $a$ when $|x|$ is large enough. Thus if the sign changes somewhere, it has to change an even number of times. Because $p$ has at most two roots, that means that the sign changes 0 or 2 times.
When $a\neq 0$, we start by solving the equation $p(x)=0$ and have three cases:


*

*$p$ has no zeros

*$p$ has one zero

*$p$ has two zeros


If $p$ has 0 or 1 zeros, its sign cannot change at any point. We can determine the sign by plugging in any value of $x$ (e.g. $x=0$). Alternatively, we may deduce it from the sign of $a$ as above. Depending on the case, we might be interested in either $p(x)\leq 0$ or $p(x)<0$ so the case of exactly 1 zero should be taken care of separately.
If $p$ has two zeros, say $x_0$ and $x_1$ with $x_0<x_1$, we can write $p(x)=a(x-x_0)(x-x_1)$. From this one can easily see that the sign of $p(x)$ is $\mathrm{sgn}(a)$ for $x<x_0$ and $x>x_1$ and $-\mathrm{sgn}(a)$ for $x_0<x<x_1$.
Doing this for $p(\pi)=MSE_\alpha(\pi) - MSE_\beta(\pi)$, we can use the results above to see how the sign behaves when $0<\pi<1$.
