The next technique I want to introduce was one of my favorites from the Math Olympiad. I think that everyone trying to qualify for the USAMO team knows this - whereas most people taking the SMO don't know about this trick. This is very useful when solving Olympiad questions involving inequalities where you can identify a function to be maximized/minimized (let's denote this $f$) and a constraint function (let's denote this $g$). This isn't well-liked by the judges, since it's an unsaid tenet of the Olympiad to solve problems through some elegant algebraic manipulation, while Lagrange multipliers essentially reduce this 'theorycrafting' stage to plain calculation. I call this brute-forcing.
Part 2: Lagrange multipliers
There's one such question in the SMO 2009 preliminary round, if I recall correctly. But my favorite illustration of this is through this problem:
"$A$, $B$, $C$ are angles in a triangle. Show that $\sin A+\sin B+\sin C \leqslant \dfrac{3\sqrt{3}}{2}$"
Observe that $A+B+C-\pi=0$. Let $f\left(A,B,C\right)=\sin A+\sin B+\sin C$ and $g\left(A,B,C\right)=A+B+C-\pi$. $f$ and $g$ have continuous first partial derivatives, and $\nabla g \neq 0$. From the theorem of Lagrange multipliers, $\nabla f=\lambda \nabla g$ where $\lambda$ is a constant. It is obvious from equating $\lambda$ that $\cos A=\cos B=\cos C$, hence $A=B=C=\dfrac{\pi}{3}$. Thereby, $f$ has an extremum at $\left(\dfrac{\pi}{3},\dfrac{\pi}{3},\dfrac{\pi}{3}\right)$, which almost closes our problem.
Now, if you were to solve this by the typical means, you will have to make use of the smoothing principle, which needs some intuition - whereas using the method above for your proof is simply mechanical calculation (some easy partial differentiation).
This technique has some applications for non-Olympiad questions too. For example,
"Find the shortest distance from the origin to the locus of $\left|z-w\right|=2$ given that $w=1+3i$ on an Argand diagram."
will be quite hard for a H2 Math question. This is essentially the same as finding the shortest distance from $\left(0,0\right)$ to $\left(x-1\right)^{2}+\left(y-3\right)^{2}=2$ on the xy-plane. They might do you a favor by giving a circle centred at $\left(1,1\right)$, $\left(2,2\right)$, ... but it's just as easy with Lagrange multipliers.
We can identify:$f\left(x,y\right)=x^{2}+y^{2}$
$g\left(x,y\right)=\left(x-1\right)^{2}+\left(y-3\right)^{2}-2$
Solving for $\nabla f=\lambda\nabla g$,
$\left(\begin{array}{c}2x\\2y\end{array}\right)=\lambda\left(\begin{array}{c}2\left[x-1\right]\\2\left[y-3\right]\end{array}\right)$
we have $\dfrac{x-1}{x}=\dfrac{y-3}{y}$ i.e. $y=3x$. Substituting,
$\left(x-1\right)^{2}+3^{2}\left(x-1\right)^{2}=2$
$10\left(x-1\right)^{2}=2$
and it's basically calculation work from here. (Be careful: there are two extrema, and we have to take the square root of the extremum value of $x^2+y^2$). This technique can also be applied to 'A'-Level questions involving vector planes, making it one of the most versatile college techniques for high school usage.
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