How can one solve this equality geometrically?

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 ?

• This plot shows roughly the situation link. We have the circle and the line, if the line is to obey the inequality, it must either not touch the circle, or be it's tangent. In the plot you can see it's the tangent. This is of course not a rigorous proof, and you have to show it's a tangent. The answer by eranreches is equivalent to what I sketched here. Commented Apr 3, 2015 at 21:56

$$\begin{gathered} x + 2y = \frac{1}{2} \hfill \\ x = \frac{1}{2} - 2y \hfill \\ {x^2} = \frac{1}{4}{(1 - 4y)^2} \hfill \\ \end{gathered}$$

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.