Show that $\log (xy) \leq \log\Big(\frac{1}{2}\Big)$

Question

I'm trying to sovle the following question:

Show that $\log (xy) \leq \log\Big(\frac{1}{2}\Big)$ if $(x, y)$ is a point of the unit circle $x^2 + y^2$ = 1 in the open first quadrant $x > 0, y > 0$.

I have used the Lagrange multiplier method to show that the point $\Big (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big)$ satisfied the constraint equation, whch yeilds the value $\log\Big(\frac{1}{2}\Big)$. I am not sure, however, how to to argue that this is the (absolute maximum) of the function $\log(xy)$ on the given curve. I don't think I can compactify the region by considering the unit circle in the first quadrant, because points on the coordinate axes have to be excluded.

Any suggestions? I'm looking for a solution which doesn't involve converting this problem to single variable problem.

Answer 1

Hint: Take exponential on both sides (why?), so you need to prove $$xy\leq \frac{1}{2}$$ It is fine to assume $(x,y) = (\cos \theta,\sin\theta)$ (why?) for some $\theta$. Now try to see if you can prove this.

Answer 2

\begin{align} &(x-y)^2\ge0\\ \implies&x^2-2xy+y^2\ge0\\ \implies&-2xy\ge-1 \end{align}

Do you know how to continue?

Answer 3

AM GM

$x^2+y^2 \ge 2\sqrt{xy}.$

Hence $1\ge 2\sqrt{xy}.$

Note: $0<x,y \le 1$, $xy \le 1.$

Then:

$\sqrt{xy} \ge xy.$

$(1/2) \ge \sqrt{xy} \ge xy$.

$\log$ is an increasing function,

hence $\log (1/2) \ge \log (xy) .$

Answer 4

ln(xy) ≤ ln(1/2) _ as ln is a bijective, increasing function, we know that: xy ≤ 1/2. making the reasonable sub: (x,y) = (cos(t), sin(t))

for it satisfies the equation: $xˆ2 +yˆ2 = 1$

We find that: $2xy ≤ 1 \rightarrow 2 \sin(t)\cos(t) ≤ 1\rightarrow \sin(2t) ≤ 1$ which is true because we know that $\sin(t)$ is bounded between 1,-1.