Fibonacci's problem

The Arithmetic mean - Harmonic Mean (AM - HM) inequality state that:

If ${ a }_{ 1 }>0, { a }_{ 2 }>0,\quad ...\quad , { a }_{ n }>0$ :

$\frac { { a }_{ 1 }+{ a }_{ 2 }+\quad ...\quad +{ a }_{ n } }{ n } \ge \frac { n }{ \frac { 1 }{ { a }_{ 1 } } +\frac { 1 }{ { a }_{ 2 } } +\quad ...\quad +\frac { 1 }{ { a }_{ n } } }$

Therefore, since $x>0, y>0, z>0$ :

$\frac { x+y+z }{ 3 } \ge \frac { 3 }{ \frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } }$

We can rearrange to get:

$\frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \ge \frac { 9 }{ x+y+z }$

Since $x+y+z = 1$ ,

$\frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \ge \frac { 9 }{ 1 }$

$\frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z } \ge 9$

Therefore, the minimum value of $\frac { 1 }{ x } +\frac { 1 }{ y } +\frac { 1 }{ z }$ is 9.

This problem can be approached by minimization-maximization logic as follows

1) To minimize the sum of (1/x) + (1/y) + (1/z) , each term should be minimized.

2) To minimize 1/x , "x" should be maximized, similarly "y" and "z" should also be maximized.

3) Since x + y + z = 1 (sum of x,y and z is equal to 1), to maximize x,y and z simultaneously, make equal values (0.33.... + 0.33.... + 0.33....) of x, y and z such that the sum is 1. If we try to maximize only one value, other values will get minimized (for e.g. 0.8 + 0.1 + 0.1). Hence, we should try to make x,y and z equal.

           x    +    y    +    z       =   1
          1/3   +   1/3   +    1/3     =   1
                               3/3     =   1

4) The maximum values are x = 1/3, y = 1/3, z = 1/3

5) Hence, the minimum value of (1/x) + (1/y) + (1/z) is

            (1/x) + (1/y) + (1/z)     =    3 + 3 + 3 
                                      =    9

First solution :

Let's note, that

$\frac { 1 }{ x } =\frac { x+y+z }{ x }=1+\frac { y }{ x } +\frac { z }{ x }$

$\frac { 1 }{ y } =\frac { x+y+z }{ y }=1+\frac { x }{ y } +\frac { z }{ y }$

$\frac { 1 }{ z } =\frac { x+y+z }{ z }=1+\frac { x }{ z } +\frac { y }{ z }$

So :

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

Let's prove that :

$a+\frac { 1 }{ a } \ge 2$ if $a>0$

It's very easy : ${ a }^{ 2 }-2a+1\ge 0$ and ${ (a-1) }^{ 2 }\ge 0$

It means, that:

$\frac { x }{ y }+\frac { y }{ x } \ge 2$

So:

$3+2+2+2=9$

Second solution :

Actually Fibonacci is not relevant to the problem.

Just look at the answer choices, you can notice that $9$ is not Fibonacci's number. So, it's the right answer =)