Fundamental Inequalities - Part 1

$\frac{3}{x + 2} + \frac{3}{y + 2} = 1$ rearranges to $y = \frac{x + 8}{x - 1}$ , therefore since $y > 0$ then $x > 1$ . Then $x + 2y = x + 2(\frac{x + 8}{x - 1}) = 3 + x - 1 + \frac{18}{x - 1}$ . By the AM-GM inequality (and since $x > 1$ ), $x - 1 + \frac{18}{x - 1}$ has a minimum at $2\sqrt{(x - 1)\frac{18}{x - 1}} = 6\sqrt{2}$ . Therefore, $x + 2y$ has a minimum at $3 + 6\sqrt{2}$ , so $a = 3$ , $b = 6$ , $c = 2$ , and $a + b + c = \boxed{11}$ .

Let $u = x+2y, v = x$ . We need to find the minimum value of $u$ . Since $x>0, y>0$ , then $u>0, v>0$ . While $x = v, 2y = u-v$ :

$\frac { 3 }{ x+2 } +\frac { 3 }{ y+2 } =1$

$\frac { 3 }{ x+2 } +\frac { 6 }{ 2y+4 } =1$

$\frac { 3 }{ v+2 } +\frac { 6 }{ u-v+4 } =1$

$\frac { 6 }{ u-v+4 } =\frac { v+2 }{ v+2 } -\frac { 3 }{ v+2 }$

$\frac { 6 }{ u-v+4 } =\frac { v-1 }{ v+2 }$

$\frac { u-v+4 }{ 6 } =\frac { v+2 }{ v-1 }$

$u=v-4+6\cdot \frac { v+2 }{ v-1 }$

$u=v-4+6+6\cdot \frac { 3 }{ v-1 }$

$u=v+2+\frac { 18 }{ v-1 }$

Let's find the derivative:

$\frac { \partial u }{ \partial v } =1-\frac { 18 }{ { \left( v-1 \right) }^{ 2 } }$

The derivative is negative when $v=0$ and positive when $v$ is a big number - there is a global minimum in range $\left[ 0,+\infty \right]$ . The minimum is reached when the derivative equals zero:

$\frac { \partial u }{ \partial v } =1-\frac { 18 }{ { \left( v-1 \right) }^{ 2 } }=0$

$1=\frac { 18 }{ { \left( v-1 \right) }^{ 2 } }$

${ \left( v-1 \right) }^{ 2 }=18$

$v-1=\pm 3\sqrt { 2 }$

$v=1\pm 3\sqrt { 2 } ,\quad v>0$

$v=1+3\sqrt { 2 }$

Then:

$u=v+2+\frac { 18 }{ v-1 } =3+6\sqrt { 2 }$

$a=3,b=6,c=2$

$a+b+c=\boxed { 11 }$

Note: The minimum is reached when: $x=1+3\sqrt { 2 }, y=1+\frac { 3\sqrt { 2 } }{ 2 }$

We can safely multiply both sides by (x+2)(y+2) to get 3(y+2)+3(x+2)=xy+2x+2y+4. Rearranging gives our first important equation shown below. $(x-1)(y-1)=9$ Now let k=x+2y and X=x-1 and Y=y-1. Our problem is asking to minimize k which can be rewritten as $k=X+2Y+3$ It is important to realize that minimizing k (equivalent to X+2Y+3) is the same as minimizing X+2Y. So if we minimize X+2Y then we can add 3 to get k and be done. From the first equation, we have XY=9 or equivalently $X*(2Y)=18$ By AM-GM we have $(X+2Y)/2\geq3\sqrt{2} \rightarrow X+2Y=6\sqrt{2}$ Remember that what we need is $k=X+2Y+3=3+6\sqrt{2}$ . So the answer is 11. Sorry about the EXTREMELY bad usage of LaTeX. Using $stuff$ and $$stuff$$ doesn't give math mode and so I had to resort to display mode. If anyone knows how to do math mode please tell me in the comments.