Given that $x,y$ are real such that: $$x+2y=\dfrac{1}{2},$$ how can one show, geometrically that $$x^2+y^2\geq \dfrac{1}{20}?$$
I see that $x^2+y^2-\dfrac{1}{20}=5\left(y-\dfrac{1}{5}\right)^2$
But in the classroom our teacher talked about an intersection between two straight lines which are perpendicular.
But I didn't understand it. Can someone tell me if that is true and why ?
On
$$\begin{gathered} x + 2y = \frac{1}{2} \hfill \\ x = \frac{1}{2} - 2y \hfill \\ {x^2} = \frac{1}{4}{(1 - 4y)^2} \hfill \\ \end{gathered} $$
Now add ${y^2}$ on both sides: $$\begin{gathered} {x^2} + {y^2} = {y^2} + \frac{1}{4}{(1 - 4y)^2} \hfill \\ = {y^2} + \frac{1}{4}(1 - 8y + 16{y^2}) \hfill \\ = 5({y^2} - \frac{2}{5}y) + \frac{1}{4} \hfill \\ = 5({y^2} - \frac{2}{5}y + \frac{1}{{25}} - \frac{1}{{25}}) + \frac{1}{4} \hfill \\ = 5{(y - \frac{1}{5})^2} - \frac{1}{5} + \frac{1}{4} \hfill \\ = 5{(y - \frac{1}{5})^2} + \frac{1}{{20}} \hfill \\ \end{gathered} $$
So we have $${x^2} + {y^2} = 5{(y - \frac{1}{5})^2} + \frac{1}{{20}}$$
Because $$5{(y - \frac{1}{5})^2} \geqslant 0$$ and $$5{(y - \frac{1}{5})^2} + \frac{1}{{20}} \geqslant \frac{1}{{20}}$$ it follows $${x^2} + {y^2} = 5{(y - \frac{1}{5})^2} + \frac{1}{{20}} \geqslant \frac{1}{{20}}$$ or $${x^2} + {y^2} \geqslant \frac{1}{{20}}$$
Notice that $d=\sqrt{(x-0)^2+(y-x)^2}=\sqrt{x^2+y^2}$ is the distance of a point $(x,y)$ from the origin.
In your question you need to show that the minimal distance between the origin and the given line is $\sqrt{\frac{1}{20}}$.
It is known that the minimal distance between a point and a line is the length of the perpendicular to the line that passes through the point, and thats proves the inequality.