Optimize over unit circle to prove $|ax + by| \le \sqrt{a^2 + b^2}$

Question

I have the following problem which, straight off the shelf, seems totally approachable. It's been giving me difficulty however:

Let $a,b,x,y \in \mathbb{R}$, and suppose that $x^2 + y^2 =1$. Prove that $|ax + by| \le \sqrt{a^2 + b^2}$.

I have started by considering the function $f(x,y) = ax + by$, and by defining a constraint function $g(x,y) = x^2 + y^2$. Then if I can maximize and minimize $f(x,y)$ subject to the constraint $g(x,y) = 1$, I should be done. By both methods (the method of Lagrange multipliers, and the method of evaluating $f(x,y)$ on the unit circle, finding critical points, etc.) I am having difficulties.

For the Lagrange multiplier method, we know that we will maximize/minimize $f(x,y)$ subject to the constraint $g(x,y) = 1$ if the system $\nabla g = \lambda \nabla f$, $g(x,y) = 1$ is satisfied. In terms of our particular example, we obtain the system of three equations

\begin{align*} 2 \lambda x &= a\\ 2 \lambda y &= b\\ x^2 + y^2 &=1. \end{align*}

I'm not sure how to solve this system. I considered using the function $h(x,y) = (f(x,y))^2 = (ax + by)^2$ instead of $f$, but then things don't work out nicely for the minimum values of $f(x,y)$ and $h(x,y)$.

For the other method, I'm even less sure of what I'm doing. Any help would be greatly appreciated.

Answer 1

First of all, there is a fairly easy way to show this inequality:

$$|ax + by| = |(a, b)\cdot (x, y)| \leq \|(a, b)\|\|(x, y)\|\cos\theta \leq \|(a, b)\|\|(x, y)\| = \|(a, b)\| = \sqrt{a^2 + b^2}.$$

As for the Lagrange multipliers method, you're almost there. As $2\lambda x = a$ and $2\lambda y = b$, $x = \frac{a}{2\lambda}$ and $y = \frac{b}{2\lambda}$. To determine the values of $\lambda$, we use the fact that $g(x, y) = 1$. Using this, we have

$$1 = g(x, y) = x^2 + y^2 = \left(\frac{a}{2\lambda}\right)^2 + \left(\frac{b}{2\lambda}\right)^2 = \frac{a^2+b^2}{4\lambda^2}$$

so $\lambda = \pm\frac{1}{2}\sqrt{a^2+b^2}$.

If $\lambda = \frac{1}{2}\sqrt{a^2+b^2}$ then $(x, y) = \left(\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}\right)$, and $f(x, y) = \sqrt{a^2+b^2}$.

If $\lambda = -\frac{1}{2}\sqrt{a^2+b^2}$ then $(x, y) = \left(-\frac{a}{\sqrt{a^2+b^2}}, -\frac{b}{\sqrt{a^2+b^2}}\right)$, and $f(x, y) = -\sqrt{a^2+b^2}$.

So $f$ has attains it maximum value of $\sqrt{a^2+b^2}$ at the first point and a minimum value of $-\sqrt{a^2+b^2}$ at the second point. Therefore, for all $(x, y)$ on the unit circle,

$$-\sqrt{a^2+b^2} \leq f(x, y) \leq \sqrt{a^2+b^2}.$$

So $|ax+by| = |f(x, y)| \leq \sqrt{a^2+b^2}$.

Answer 2

For the other method, parametrize the circle by $x=\cos t$ and $y=\sin t$. Then $$ ax+by=a\cos t+ b\sin t=\sqrt{a^2+b^2}\sin(t+\phi) $$ for some $\phi$ (you should be able to check this). Maybe you can continue from here?

Answer 3

can we do this. we will show that you can write $$ ax + by = \sqrt{a^2 + b^2}\cos(t - \phi)$$ which will give us the inequality $$ |ax + by| \le \sqrt{a^2 + b^2}$$ here is an outline:

since $x^2 + y^2 = 1,$ there is $t$ such that $x = \cos t, y = \sin t$ and $ax + by = \sqrt{a^2 + b^2}\left(\dfrac{a}{\sqrt{a^2+b^2}} \cos t + \dfrac{b}{\sqrt{a^2+b^2}} \sin t\right) = \sqrt{a^2 + b^2}\cos(t-\phi)$ where $\phi$ is determined by $ \cos \phi = \dfrac{a}{\sqrt{a^2+b^2}}, \ \sin \phi = \dfrac{a}{\sqrt{a^2+b^2}}$

Answer 4

working backwards gives by squaring $$a^2x^2+b^2y^2+2abxy\le a^2+b^2$$ this is equivalent to $$2abxy\le a^2(1-x^2)+b^2(1-y^2)$$ or $$2abxy\le a^2y^2+b^2x^2$$ and this is $$0\le (ay-bx)^2$$ which is true.

Answer 5

Your approach is reasonable.

Let $f((x,y)) = |ax+by|$, $g((x,y)) = {1 \over 2}(x^2+y^2)$, solve $\max \{f((x,y))| g((x,y)) = {1 \over 2} \}$.

The constraint $g((x,y)) = {1 \over 2} $ shows that the feasible set is compact, so we know a maximum exists.

Note that if $a=b=0$, then the result is true, so suppose $(a,b) \neq 0$.

If $(a,b) \neq 0$, choosing $(x,y) = {1 \over \sqrt{a^2+b^2}} (a,b)$ shows that $ax+by = \sqrt{a^2+b^2} >0$, so the problem is equivalent to solving $\max \{ ax+by| g((x,y)) = {1 \over 2} \}$. The point here is that I can remove the $|\cdot|$.

Noting that the gradient of the constraint is never zero, we use Lagrange multipliers to get $\begin{bmatrix} a \\ b \end{bmatrix} + \lambda \begin{bmatrix} x \\ y \end{bmatrix} = 0$. Note that $\lambda \neq 0$ since $(a,b) \neq 0$.

This gives $x = { a \over \lambda }$, $y = { b \over \lambda }$, and using the constraint $g((x,y)) = {1 \over 2}$ shows that $\lambda = \pm\sqrt{a^2+b^2}$. Checking shows that both values of $\lambda$ are solutions.

Hence the solutions are $(x,y) = \pm{1 \over \sqrt{a^2+b^2}} (a,b)$.

Alternative approach:

The problem can be written in the form $\max \{ |\langle c , x \rangle | \mid \|x\| = 1 \}$, where $c,x \in \mathbb{R}^n$.

We can write any $x= \alpha {c \over \|c\|} + v$, where $v \bot c$, and it is easy to check that $\|x\|^2 = \alpha^2 + \|v\|^2$. Hence we can write the problem as $\max \{ |\alpha| \|c\| \mid \alpha^2 + \|v\|^2 = 1, v \bot c \} $. It is straightforward to see that $\alpha = 1, v=0$ solves the problem, so the maximum is $\|c\|$ and is attained at $x={c \over \|c\|}$.

(Note: This proves the Cauchy-Schwarz inequality on $\mathbb{R}^n$ since we have $|\langle c,d \rangle| = \|d\| |\langle c,{d \over \|d\|} \rangle| \le \|d\| \|c\|$.)