Show that $\log (xy) \leq \log\Big(\frac{1}{2}\Big)$ 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 $x &gt; 0, y &gt; 0$.


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

Do you know how to continue?
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.