Show that if $2x + 4y = 1$ where x and y are real numbers.

Question

Show that if $2x + 4y = 1$ where $$x, y \in \mathbb R $$, then $$x^2+y^2\ge \frac{1}{20}$$

I did this exercise using the Cauchy inequality, I do not know if I did it correctly, so I decided to publish it to see my mistakes. Thank you!

If $x=2$ and $y=4$, then $$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$ iff $$\frac{x}{2}=\frac{y}{4}$$, then $$20\ge(2x+4y)^2$$ $$-4.47\le2x+4y\le4.47$$

Answer 1

You have one big confusion here. It is always true:$$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$ with equality iff $\frac{x}{2}=\frac{y}{4}$. Now, since $2x+4y=1$ we get $$20(x^2+y^2)\ge 1$$ so $$x^2+y^2\ge \frac{1}{20}$$ and you are done.

Now if you are interested when eqaulity ocurres then just plug $y=2x$ in $2x+4y=1$ and you get $y=1/10$ and $x=1/5$.

Answer 2

Your work is correct up to $$(2^2+4^2)(x^2+y^2)\ge (2x+4y)^2$$

From which you get $$ 20(x^2+y^2)\ge (1)^2$$

Therefore, $$x^2+y^2\ge \frac{1}{20}$$

At this point your are done.

Answer 3

The distance from the circle's center ($ x^2+y^2 = \frac{1}{20}$) to the line $2x + 4y = 1$ is given by

$$ d = \frac{1}{\sqrt{2^2+4^2}} = \sqrt{\frac{1}{20}} $$

Answer 4

It suffices to show that the line $2x+4y=1$ crosses the circle $x^2+y^2=1/20$ in at most one point.

I made a few substitutions and got $$25y^2-10y+1=0,$$ which clearly has exactly one solution. This proves the result.