How do I make this simple proof better (and more correct?) Let $x$ and $y$ be real numbers. If $x\cdot{y}&gt;\frac{1}{2}$, then $x^2+y^2&gt;1$.

Proof: We will prove with the direct method. Let $x$ and $y$ be real numbers. Since
$$ x\cdot{y}&gt;\frac{1}{2} $$
it follows that
$$ 2xy&gt;1,$$
which means
$$x^2+y^2 \geq 2xy.$$
Therefore,
$$ x^2+y^2&gt;1. $$
Write $0\le (x-y)^2=x^2-2xy+y^2$, hence $1&lt;2xy\le x^2+y^2$.
There is a much simpler proof.

Since $xy &gt; \frac{1}{2}$, $x$ and $y$ must be non zero numbers, which are both positive or negative.

Suppose $x&gt;0$ and $y&gt;0$.

If $x&gt;1$, then also $x^2&gt;1$, and $x^2 +y^2&gt;x^2&gt;1$ is obvious.

Take $0&lt;x\leq 1$ and $y&gt;0$.

Since $0&lt;x\leq 1$, the function $f$ with $f(x)=\sqrt{1-x^2}$   is well defined.  Let  $g$ be the function with $g(x)=\frac{1}{2x}, \ 0&lt;x\leq 1$.

It can be easily proved that the function h with $h(x)=-x+\sqrt{2}, \ 0&lt;x\leq 1$ 
      has a graph which is a line tangent to the graphs of $f$ and $g$ at the point
      $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$.

One can also verify easily that $f$ is a concave function and that $g$ is a convex funtion. Therefore the tangent line $h$ is above the graph of $f$ and below the graph of $g$. 
      So if  $0&lt;x\leq 1$,the previous remark implies that  $f(x)\leq h(x)\leq g(x)$ and  $\sqrt{1-x^2}\leq \frac{1}{2x}$.

Now, the proposition we want to prove is obvious: If $xy &gt; \frac{1}{2}$, then $y&gt;\frac{1}{2x}$, 
      and $\frac{1}{2x} \geq \sqrt{1-x^2}$ from the previous inequality, hence 
      $y&gt; \frac{1}{2x} \geq \sqrt{1-x^2}$. Consequently, $y&gt;\sqrt{1-x^2}$, which implies
      $y^2&gt;1-x^2$ and therefore $x^2+y^2&gt;1.$

The case $x&lt;0$ and $y&lt;0$ is similar.