Strange proof of Schwarz Inequality with Pythagorean Theorem Does anyone know what is going on in t

Strange proof of Schwarz Inequality with Pythagorean Theorem
Does anyone know what is going on in this proof of the Schwarz inequality? Most importantly: how can one assume that $c^2\leqq <!--\; \leqq \; -->\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$, or later on, that $c^2\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->\leqq <!--\; \leqq \; -->\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$? This would imply that $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cE{\Vert <!--\; \Vert \; -->}^2$, or in the latter case $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cB{\Vert <!--\; \Vert \; -->}^2$, could also be equal to zero (seeing as it's a smaller-than-or-equal-to sign). But since we're using the Pythagorean theorem to arrive at these relations, $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cE{\Vert <!--\; \Vert \; -->}^2$ or $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->cB{\Vert <!--\; \Vert \; -->}^2$ could not possibly be zero, as then one of the sides of our 'triangle' would be zero, and we wouldn't be able to use the Pythagorean theorem to justify this conclusion. So why isn't it just a '<'-sign? As I said, I don't see how vector A−cE or A−cB can have length zero if we're using the Pythagorean theorem, that is, if we're presupposing the existence of a triangle. If either A−cE or A−cB equals zero, there wouldn't be a triangle anymore, and we wouldn't be able to apply the Pythagorean theorem.
Edit: fixed link.
Edit2: I must add that the author maintains an unusual definition for the scalar component; instead of the conventional $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\mathbf{B}}{\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->}$, i.e., $\mathbf{A}\cdot <!--\; \cdot \; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{B}}$, the author defines $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\mathbf{B}}{\mathbf{B}\cdot <!--\; \cdot \; -->\mathbf{B}}$ to be the scalar component, i.e., $\frac{\mathbf{A}\cdot <!--\; \cdot \; -->\stackrel{\mathbf{^<!--\; ^\; -->}}{\mathbf{B}}}{\Vert <!--\; \Vert \; -->B\Vert <!--\; \Vert \; -->}$. In other words, the author divides the 'conventional' scalar component by the the norm of B, thus making 'his' scalar component a number which is in fact the proportion of the 'regular' scalar component to the entire length of the vector on which this component is projected.
Edit3: Since people don't seem to understand my confusion, let me phrase my question as explicitely as possible. My question is: why is the $\ge <!--\; \ge \; -->$ sign used? Why not just ">"? In what situation could $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$ equal $t^2\Vert <!--\; \Vert \; -->E{\Vert <!--\; \Vert \; -->}^2$ (the $\ge <!--\; \ge \; -->$ sign means greater than OR EQUAL TO, so in what situation could the left hand side ever equal the right hand side?) The problem is: since we're using the Pythagorean theorem for our proof, we're presupposing the existence of a triangle (or otherwise we wouldn't be able to use the Pythagorean theorem), and as such, $\Vert <!--\; \Vert \; -->A{\Vert <!--\; \Vert \; -->}^2$ can never equal $t^2\Vert <!--\; \Vert \; -->E{\Vert <!--\; \Vert \; -->}^2$, because this would mean that $\Vert <!--\; \Vert \; -->A-<!--\; -\; -->tE{\Vert <!--\; \Vert \; -->}^2$ is equal to zero, and that one of the sides of our triangle has length zero. Then we wouldn't have a triangle, and we wouldn't be justified in using the Pythagorean theorem in our particular situation. So let me ask it again: why is the $\ge <!--\; \ge \; -->$ sign used, instead of the ">"-sign?