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.
 A: 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.
A: 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 $x^2+y^2<1$.
A: 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 $1+0$ $(=1)$.
A: 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.
A: 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, $D=6/\sqrt{13}\approx 1.6641$.
A: 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.
A: Here are two geometric ways to prove this:
Method.$(1)$
Let the line intersect with the x-axis at point $A(2,0)$, and intersect with the y-axis at point $B(0,3)$. Draw the height from  $O$ to the hypotenuse and intersect with $AB$ at $H$, $\Rightarrow OH\perp AB$. So $OH$ is the shortest distance from $O$ to the straight line $AB$.
So we have $OA=2,~OB=3,~AB=\sqrt{13}$. From the triangle area, we have:
$$\frac{1}{2}\cdot OA\cdot OB=\frac{1}{2}\cdot AB\cdot OH~~\Rightarrow~~OH=\frac{6}{\sqrt{13}}$$
Because $OH$ is the shortest distance from $O$ to the straight line $AB$, and $OH>1=R$, where the radius of the circle is $R=1$. Therefore, the straight line $AB$ is outside the circle. So for all points on the straight line, which satisfying $\frac{x}2+\frac{y}3=1$, their distances to the origin $O$ are greater than $1$. Namely, $x^2+y^2>1$.
Method.$(2)$
Combine two equations:
$$\frac{x}2+\frac{y}3=1,~~~x^2+y^2=1$$
Eliminating $x$ (or eliminating $y$), and we get a quadratic equation, it is easy to show it has no real roots, which means there are no intersecting points or touching points between the straight line and the circle. This means the line is outside the circle. So for all points on the straight line, which satisfying $\frac{x}2+\frac{y}3=1$, their distances to the origin $O$ are greater than $1$. Namely, $x^2+y^2>1$.
