# Suppose that $x$ and $y$ satisfy $\frac{x}{2} + \frac{y}{3} = 1$. Prove that $x^2 + y^2 > 1$.

Ok , i tried to prove this via Contrapositive setting $x^2 + y^2 \le 1$. After doing some algebra i have arrived at $x \le \sqrt{-y^2}$. I'm fairly sure this isn't right. I also solved for x and y in equation one hoping this would somehow lead me to a conclusion, it didn't.

• Where did the $1$ disappear? – Berci Mar 9 '14 at 23:54
• I re wrote the right hand side as sqrt(1) - sqrt(-y^2) so the sqrt(1) just becomes 1. Hence sqrt(-y^2) remains? – Achilles Mar 9 '14 at 23:58
• If it were $x\le\sqrt{1-y^2}$, would make more sense.. – Berci Mar 10 '14 at 0:00

While the above answers are correct, there's no need for anything as complicated as what's been posted so far. Given the (unstated) assumption that $x$ and $y$ are both real numbers, then $x^2\geq0$ and likewise $y^2\geq0$. Now assume provisionally that $x=y=1$. Then $\frac{x}2 +\frac{y}3=\frac{5}6$, which is less than $1$. So at least one of the two variables must be greater than $1$, and likewise its square must be greater than $1$. Since we already know that the square of the other variable must be at least $0$, it follows that the sum $x^2 + y^2$ must be greater than $1+0$ $(=1)$.

Your line is, after multiplying by 6 and moving things to the left, $3x+2y-6=0$, whose distance to the origin is $|-6|/\sqrt{3^2+2^2}=6/\sqrt{13}\approx 1.6641.$

See this page for the distance from point to line formula, with several proofs of it.

Added: to proceed via the contrapositive, assume in fact that $x^2+y^2 \le 1$. Then we have $|x|\le 1$ and $|y| \le 1$, from which $$3x+2y \le 3|x|+2|y| \le 3+2=5,$$ making $3x+2y=6$ not possible.

• I am trying to prove it via the contrapositive method though. – Achilles Mar 10 '14 at 0:04
• @Achilles OK I'll think about that. – coffeemath Mar 10 '14 at 0:05
• Thank you , ill be sure to comment back immediately if i figure it out. – Achilles Mar 10 '14 at 0:08
• I see makes good sense now , just one quick question if i may. What allows us to say |x| <= 1 and |y|<= 1? – Achilles Mar 10 '14 at 0:19
• @Achilles If $|x|>1$ then also $x^2>1$ so that, since $y^2 \ge 0$, we would have $x^2+y^2>1$ but we are assuming the sum of the squares is at most $1$ in this instance. – coffeemath Mar 10 '14 at 0:23

Hint: Try the other way: $x/2+y/3=1$, so, $\frac32x+y=3$, yielding $y=3-\frac32 x$. Then compute $x^2+y^2$.

Or, even better: try geometrically: the set of points $(x,y)$ on the plane that satisfy $x/2+y/3=1$ is a line. This line contains $(2,0)$ and $(0,3)$. Draw it and draw also the disk $x^2+y^2<1$.

• Ok, i came up with the y = 3 - 3/2x. Now your just saying sub it in for y and do the algebra? – Achilles Mar 10 '14 at 0:01
• yes, that was the first suggestion. – Berci Mar 10 '14 at 0:02

You have $$x^2 + (3 - 3/2x)^2 = 9 - 9x + {13\over 4}x^2 = {13\over 4}\left(x^2 - {36\over 13}x + \right)+9= {13\over 4}\left(x^2 - {36\over 13}x + {324\over 169}\right)+ 9 - {13\over 4}\cdot {324\over 169}$$

This expression evaluates to $${13\over 4}\left(x^2 - {36\over 13}x + {324\over 169}\right)+ 9 - {81\over 13} = {13\over 4}\left(x^2 - {18\over 13}\right)^2 + {36\over 13}$$

There is a lower bound of $36/13$, which is larger than 1.

• I see , quite the rigorous proof here, thank you. The After seeing this it shows me the power of the contrapostive method. – Achilles Mar 10 '14 at 0:23

Draw the standard coordinate axes with your line. Now consider the triangle formed by the intersection of these axes with the line in the first quadrant. (A right triangle with base length=2 and height=3.)

Now draw a new line from the origin to your line such that the lines are perpendicular. This point of intersection is the closest your line ever gets to the origin, and thus serves as a lower bound for $\sqrt{x^2+y^2}$. Let us use the law of sines on the new triangle we just formed to find the distance $D$: $$\frac{\sin\arctan(3/2)}{D} = \frac{\sin\left(\pi/2)\right)}{2}.$$

As was observed by @coffeemath, $D=6/\sqrt{13}\approx 1.6641$.

Just to add variety to the answers, we can prove directly using the Cauchy-Schwarz inequality:

$$(x^2+y^2)\left(\frac{1}{2^2} + \frac{1}{3^2}\right) \ge \left(\frac{x}{2} + \frac{y}{3}\right)^2$$ Manipulating, we have $$x^2+y^2 \ge \frac{36}{13}$$ $$x^2 + y^2 > 1$$

Of course, this approach assumes that the Cauchy-Schwarz inequality is true.