How to prove the following inequality $x+y\ge2$

Question

Let $x$ and $y$ be two real positive integers, such that: $x+y+xy=3$ prove that $x+y\ge2$ I tried some simplifications like this one $x(1+y)=3-y$ and $y(1+x)=3-x$ and using the fact that both of $x$ and $y$ are positive may help us but I can't solve the problem.

Answer 1

By AM-GM $$3=x+y+xy\leq x+y+\left(\frac{x+y}{2}\right)^2,$$ which gives $$(x+y)^2+4(x+y)\geq12$$ or $$(x+y+2)^2\geq16$$ or $$x+y\geq2.$$

Answer 2

This identity needs no introduction: $$(x+y)^2=x^2 +y^2 + 2xy$$ By hypothesis: $$6=2x+2y+2xy$$ Subtracting: $$(x+y)^2 -6 = x^2 - 2x + y^2 -2y$$ Rearranging: $$(x+y)^2= (x-1)^2 + (y-1)^2 +4$$ Each term on the right hand side is non-negative, so $$(x+y)^2 \ge 4.$$ Since $x$ and $y$ are both positive: $$x+y\ge 2.$$

Answer 3

Given: $$x+y+xy=3$$ Its easy to see: $$x+y+xy+1=3+1$$ which gives, $$(1+x)(1+y)=4$$ where $(1+x),(1+y) > 0 $

Now applying AM-GM inequality: $$\frac{(1+x)+(1+y)}2 \ge \sqrt{(1+x)(1+y)} $$ which gives, $$\frac{2 + x +y}2 \ge \sqrt{4} $$ or $$x+y\geq2.$$

Answer 4

$x+y \geq 2$ doesn't need to be proved from your relation as both $x$ and $y$ are positive integers they are both greater or equal to one and thus the sum of the two must be greater or equal than two.

The fact that $4 = (1+x)(1+y)$ also implies that $x=y=1$

Answer 5

We can also take a geometrical interpretation. The equation $ \ x + y + xy = 3 \ $ describes a translated rectangular hyperbola, of which we are only interested in the portion in the first quadrant. We can rotate the curve to place its focal axis on a "horizontal" axis using the coordinate substitution $ \ x \ = \ \frac{\sqrt2}{2}·(u + v) \ , \ y \ = \ \frac{\sqrt2}{2}·(u - v) \ \ ; $ this produces $$ \frac{\sqrt2}{2}·(u + v + u - v) \ + \ \frac12·(u + v)·(u - v) \ \ = \ \ 3 \ \ \Rightarrow \ \ \sqrt2·u \ + \ \frac12·(u^2 - v^2) \ = \ 3 $$ $$ \Rightarrow \ \ \frac{u^2 \ + \ 2\sqrt2·u \ + \ 1}{2} \ - \ \frac{v^2}{2} \ \ = \ \ 3 \ + \ 1 \ \ \Rightarrow \ \ \frac{(u \ + \ \sqrt2)^2 }{2} \ - \ \frac{v^2}{2} \ \ = \ \ 4 \ \ . $$ The vertices of the hyperbola lie on the $ \ u-$axis $ \ ( \ v = 0 \ ) \ $ at $ \ (u \ + \ \sqrt2)^2 \ = \ 2·4 \ \Rightarrow \ u + \sqrt2 \ = \ \pm \ 2\sqrt2 \ \ . $ We want the vertex with positive $ \ u-$coordinate, which is then found on the line $$ x + y \ \ = \ \ \frac{\sqrt2}{2}·(u + v + u - v) \ \ = \ \ \sqrt2·u \ \ = \ \ \sqrt2·\sqrt2 \ \ = \ \ 2 \ \ . $$ All other points on the "positive branch" of the hyperbola lie at this value or larger values of $ \ u \ \ , $ so we have $ \ \sqrt2·u \ \ = \ \ x + y \ \ \ge \ \ 2 \ \ . $

The restriction to positive integers is actually excessive. Since the intercepts of the positive branch are $ \ (0 \ , \ 3) \ $ and $ \ (3 \ , \ 0) \ \ , $ the only permissible values for $ \ x \ $ would be $ \ 1 \ $ and $ \ 2 \ \ . $ We find $$ 1 \ + \ y \ + \ 1·y \ \ = \ \ 3 \ \ \Rightarrow \ \ y \ = \ 1 \ \ \ \text{and} \ \ \ 2 \ + \ y \ + \ 2·y \ \ = \ \ 3 \ \ \Rightarrow \ \ y \ = \ \frac13 \ \ . $$

So the sole positive integer solution $ \ (1 \ , \ 1) \ $ gives us $ \ x + y \ = \ 2 \ \ , $ thereby "demonstrating" the inequality.