<s>[INST] Optimize over unit circle to prove $|ax + by| \le \sqrt{a^2 + b^2}$ 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.  
 [/INST] 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 $|ax+by| = |f(x, y)| \leq \sqrt{a^2+b^2}$.
</s>