Revenge of Analysis: Using Lagrange Multipliers to Destroy Olympiad Inequalities

Goal: learn how to use Lagrange multipliers in olympiads.

Lagrange multipliers are a nice tool to solve inequalities, but they are rarely seen in olympiad solutions. Once you are comfortable with it, it can be an overpowered way smash open inequalities without much insight. (So economists love them!) For this reason, we need to approach it rigorously, to ensure we can justify earning marks.

I really recommend watching this video to get some intuition first.

All of this is covered in my course notes for IB Analysis and Topology, in the relevant sections. The results and definitions in this article are correct only for $\mathbb{R}^n$ (e.g. compactness is something different, but the Heine-Borel theorem says that in $\mathbb{R}^n$ it's equivalent to being closed and bounded). For the more general definitions and a deeper understanding, check out the course notes.

Defn.

Consider a set $M$ together with a function $d : M \times M \to \mathbb{R}$.

$(M, d)$ is a metric space if:

• $d(x,y) \geq 0$, equality if and only if ("iff") $x=y$ ("positive semi-definite")
• $d(x,y) = d(y,x)$ ("symmetric")
• $d(x,y) + d(y,z) \geq d(x,z)$ ("triangle inequality")

Defn. (balls)

The open ball in $\mathbb{R}^n$ with centre $p \in \mathbb{R^n}$ and radius $r$ is

$$B(p,r) := \{x \in \mathbb{\mathbb{R}^n} \mid d(p,x) \lt r \}$$

Similarly, the closed ball $B[p,r]$ is¹ the set:

$$\{x \in \mathbb{R}^n \mid d(p,x) \leq r\}$$

Defn.

$U \subseteq \mathbb{R}^n$ is open iff for every $p \in U$, $\exists r\gt 0$ s.t. $B(p,r) \subset U$.

Informally: an 'open set' is a set such that for any point $x$ in the set, we can fit a ball ball around $x$, while staying inside the set.

Defn. (limits)

Let $(x_k)_{k=1}^\infty$ be a sequence in $\mathbb{R}^n$.

The sequence converges to the point $x_\infty$ iff $\forall \epsilon \gt 0, \exists n_0 \in \mathbb{N}$ such that:

$$n \geq n_0 \implies d(x_n, x_\infty) \lt \epsilon$$

Then $x_\infty$ is denoted $\lim_{n \to \infty}(x_n)$.

Defn.

Let $S \subseteq \mathbb{R}^n$. $S$ is closed iff for every sequence of points $(x_k)_{k=1}^\infty$ that satisfies $x_k \in S \;\forall\; k$, we have $\left( \lim_{k \to \infty} x_k \right) \in S$.

Defn.

Let $A \in \mathbb{R}^n$. The closure of $A$, denoted $\bar A$, is the smallest closed set containing $A$.

Fact.

$A \subseteq \mathbb{R}^n$ is closed if and only if $\mathbb{R}^n \setminus A$ is open.

Fact.

Let $U,V$ be open sets. Then $U \cap V$ and $U \cup V$ are also open sets. This extends to finite intersections² and infinite unions.

Fact.

Let $S,T$ be closed sets. Then $S \cap T$ and $S \cup T$ are also closed sets. This extends to infinite intersections and finite unions.

Defn.

$A \subseteq \mathbb{R}^n$ is bounded iff $\exists\, R \in \mathbb{R}, R\gt 0$ such that $A \subseteq B(0, R)$.

Defn.

A subset $K \subseteq \mathbb{R}^n$ is compact if it is closed and bounded.

Defn.

Let $D \subseteq \mathbb{R}^n$ and let $f : D \to \mathbb{R}$.

$f$ is continuous at the point $p \in D$ iff $\forall \epsilon \gt 0$, $\exists \delta \gt 0$ such that $\forall x \in D$ we have:

$$d(p,x) \lt \delta \implies \lvert f(x) - f(p) \rvert \lt \epsilon$$

$f$ is continuous iff it is continuous at every point.

Informally, no matter how small $\epsilon$ you pick, I can always find a region around $p$ where the change in $f$ is smaller than $\epsilon$. So, a small change in input causes a small change in ouput.

Theorem. (extreme value thm)

Let $f : K \to \mathbb{R}$ be a continuous function, where $K \subseteq \mathbb{R}^n$ is a nonempty compact set.

Then $f$ has both a global maximum value and a global minimum value³:

$$\exists\, x \in K \text{ s.t. } f(x) \geq f(y) \;\forall\, y \in K$$$$\exists\, x' \in K \text{ s.t. } f(x') \leq f(y) \;\forall\, y \in K$$

Example.

Let $K$ be a closed ball in $\mathbb{R}^2$, then $K$ is compact. Let $f : K \to \mathbb{R}, f(x) = d(x,(0,0))$ which is continuous.

Then the theorem says that there is a point(s) on $K$ which is closest to $(0,0)$, and a point(s) which is furthest.

Note: We need to assume $K$ is closed for this theorem, else we can construct a counterexample where $f$ increases to infinity the closer you get to the edge.

Fact.

Let $g : \mathbb{R}^n \to \mathbb{R}$ be continuous. Then for a fixed $c \in \mathbb{R}$, the set

$$\{ x \in \mathbb{R}^n \mid g(x) = c \}$$

is closed in $\mathbb{R}^n$.

I'm assuming you've already met these, so I'll recap.

Finally!

Theorem. (Lagrange Multipliers, "LM")

Let $U \subset \mathbb{R}^n$ be an open set⁴ and let $f,g : U \to \mathbb{R}$ be continuous functions with continuous partial derivatives of the first order.

Let $c \in \mathbb{R}$ and $S = \{x \in U \mid g(x) = c\}$. (Note: $S$ doesn't have to be open or closed, that's $U$!)

Then, if $x_0 \in S$ is a local max or min, then either:

$$\left( \frac{\delta g}{\delta x}, \frac{\delta g}{\delta y}, \frac{\delta g}{\delta z},\cdots\right) = (0,0,0,\cdots)$$

Or $\exists \lambda \in \mathbb{R}$ such that:

$$\frac{\delta f}{\delta x}(x_0) = \lambda \frac{\delta g}{\delta x}(x_0),$$$$\frac{\delta f}{\delta y}(x_0) = \lambda \frac{\delta g}{\delta y}(x_0),$$$$\frac{\delta f}{\delta z}(x_0) = \lambda \frac{\delta g}{\delta y}(x_0),$$$$\text{etc.}$$

Let $x,y,z \geq 0$ such that $x+y+z = 1$. Find the min and max of $xyz$.

Solution.

Let $f(x,y,z) = xyz$ and $g(x,y,z)=x+y+z$; these are polynomial functions and so are continuous. We are maximizing and minimizing $f$, subject to a condition on $g$.

$0 \leq x,y,z \leq 1$ so we're only interested in the cube $[0,1] \times [0,1] \times [0,1]$.

$x+y+z=1$ is a plane (coloured green)

Let $U = (0,1)^3$, then $\bar U = [0,1]^3$.

Let $S = \{x \in U \mid g(x)=1\}$, then $\bar S$ is bounded hence compact.

Hence $f$ has a global max and min in $\bar S$.

The global extrema might be on the boundary of $\bar S$. If so then we cannot apply LM, because the extrema will not be in $S$.

• If we are on the boundary, then one of $x,y,z$ is $0$, so $f(x,y,z) = xyz = 0$. Thus $f$ is zero everywhere on the boundary, so $0$ would be an extremum.

• If we are not on the boundary, then we are in $S$, so we can apply LM.

$$\frac{\delta g}{\delta x} = 1, \frac{\delta g}{\delta y} = 1, \frac{\delta g}{\delta z} = 1$$

So we are in the second case of the theorem, because $(1,1,1) \neq (0,0,0)$.

$$\frac{\delta f}{\delta x} = yz, \frac{\delta f}{\delta y} = xz, \frac{\delta f}{\delta z} = xy$$

So $yz = \lambda \cdot 1$, $xz = \lambda \cdot 1$, $xy = \lambda \cdot 1$.
This implies $xy = yz = zx$ so $x=y=z$, and finally $x+y+z = 1$ $\implies$ $x=y=z=\frac{1}{3}$, so an extremal value of $f$ is $\frac{1}{27}$.

Overall, all extreme values of $f$ on $\bar S$ are $0$ or $\frac{1}{27}$.

$$\therefore 0 \leq xyz \leq \frac{1}{27}$$

Let $x,y,z \geq 0$ such that $x+y+z = 1$. Show that$$0 \leq yz+zx+xy-2xyz \leq \frac{7}{27}$$

Given that $x,y \in \mathbb{R}$ with $x^2 + y^2 = 1$, Find the max and min values of $8x^2 - 2y$.

Note: the "normal" way to do this would be to write it as $8(1-y^2)-2y$ and bound this quadratic. But we can do it with LM too. I'll let you decide which way is easier.

Suppose we want to prove some inequality, but there are no constraints.

If the inequality is homogenous, then we can impose a condition e.g. $a+b+c=1$ or $abc = 1$ or $a^2+b^2+c^2=1$, because we can scale each variable to make the condition true.

For $x,y \in \mathbb{R}, (x,y) \neq (0,0)$, prove that:$$\frac{x+y}{x^2-xy+y^2} \leq \frac{2\sqrt 2}{\sqrt {x^2 + y^2}}$$

(Note: much easier to fix $y$ and use normal derivatives, but we want to solve with LM)

Reveal spoiler

Solution.

The inequality is homogenous, so we can impose the condition $x^2 + y^2 = 1$ by scaling the variables. Thus it is sufficient to show that $\frac{x+y}{x^2-xy+y^2} \leq 2\sqrt 2$ when $x^2 + y^2 = 1$.

$$\text{Goal: } \frac{x+y}{x^2-xy+y^2} \leq 2\sqrt 2$$$$\iff 0 \leq 2\sqrt2(x^2-xy+y^2) -x - y$$$$\iff 0 \leq 2\sqrt2(1-xy) -x - y$$

Note that we didn't have to clear the $x^2 + y^2$ term, but it makes the computation easier - always look for tricks!

Now as usual, define:

$$f,g: \mathbb{R}^2 \to \mathbb{R}$$$$f(x,y) = 2\sqrt2(1-xy)-x-y$$$$g(x,y) = x^2 + y^2$$$$S = \{x \in \mathbb{R}^2 \mid g(x,y) = 1\}$$

Then $S$ is closed. Since $S$ is also bounded, we have that $S$ is compact.

Also $f,g$ are continuous with continuous partial derivatives, since they are polynomials.

Hence $f$ has a global max and min on $S$.

$\nabla g = (2x,2y) \neq (0,0)$ since $a^2 + b^2 = 1$.

$$\therefore \nabla f = \lambda \nabla g$$$$\implies \begin{bmatrix} -2\sqrt2 y - 1 \\ -2\sqrt2 x - 1\end{bmatrix} = \lambda \begin{bmatrix} 2x \\ 2y\end{bmatrix}$$

Hence,

$$-2y\sqrt2 - 1 = 2x\lambda$$$$-2x\sqrt2 - 1 = 2y\lambda$$$$x^2 + y^2 = 1$$

Solving these to find $f(x,y)$:

$$y(-2y\sqrt2 - 1) = 2xy\lambda$$$$x(-2x\sqrt2 - 1) = 2yx\lambda$$$$\therefore -2y^2\sqrt2 - y = -2x^2\sqrt2 - x$$$$\implies 2\sqrt2(x^2 - y^2) + (x-y) = 0$$

Thus either $x=y$ or $2\sqrt2(x+y) + 1 = 0$.

• If $x = y$:
  Then, since $x^2 + y^2 = 1$, we have that $x = y = \pm \frac{1}{\sqrt2}$. In this case:

$$f(x,y) = f\left(\pm \frac{1}{\sqrt2}, \pm \frac{1}{\sqrt2}\right)$$$$= 2\sqrt2(1 - \frac{1}{2}) - (\pm \sqrt 2)$$$$= \sqrt2 \mp \sqrt2$$$$= 0, 2\sqrt2$$

• If $2\sqrt2(x+y) + 1 = 0$:
  Then $x + y = -\frac{1}{2\sqrt2}$.

$$\implies (x+y)^2 = \frac{1}{8}$$$$\implies 2xy = \frac{1}{8} - (x^2+y^2) = \frac{1}{8} - 1$$$$\implies xy = -\frac{7}{16}$$

So in this case:

$$f(x,y) = 2\sqrt2(1-xy)-(x+y)$$$$= 2\sqrt2\left(1+ \frac{7}{16}\right) + \frac{1}{2\sqrt2}$$$$= \frac{25\sqrt2}{8}$$

Overall, the possible extremal values of $f$ are $0, 2\sqrt2, \frac{25\sqrt2}{8}$.

$$\therefore 0 \leq f(x,y) \leq \frac{25\sqrt2}{8}$$

In particular, $f(x,y) \geq 0$ as required.

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Let $a,b,c \gt 0$ such that $a+b+c=3$. Find the minimum value of:$$f(a,b,c) = \frac{2-a^3}{a} + \frac{2-b^3}{b} + \frac{2-c^3}{c}$$

If we attempt to use LM:

The problem is that $f$ is not defined on the boundary, so we cannot say $f$ has a global max and min in the area we're looking at (indeed $f$ can be arbitrarily large if we let $a$ approach zero for example). Boo, we can't use LM.

Actually, in this specific case we can fix it with the following approach:

Reveal spoiler

Make the triangle a bit smaller on all sides, then it is compact and we can use LM. For the region of that we didn't consider, we can assume that "one of the variables is at least this small", so "$f$ is at least this large", and get it to be larger than a value we already know.

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As you've seen, it takes careful consideration and background knowledge to use Lagrange multipliers in olympiads correctly. Best of luck!