Find the min value of $\frac{1}{x+\frac{1}{y+\frac{1}{z}}}$, if $x\ne y \ne z$ and $x,y,z\in {1,2,3,4,5}$ My answer is $\frac{5}{29}$, I just use logic to substitute numbers in the expression, but I can't prove my answer If this expression be minimum then the denominator should be the greatest, so I just let  x=5 and I need to let $\frac{1}{y+\frac{1}{z}}$ be the greatest so just need this denominator be smallest so y=1 and z=4, this is what I thought. 
 A: Note that with $x,y,z \gt 0$ you have $$x+\frac 1{y+\frac 1z}\lt x+1$$ so if $x\le 4$ the denominator will be smaller than if $x=5$ and the fraction larger. You can repeat this kind of argument to make a proof (you will want $y$ to be small and $z$ large for similar reasons), and generalise it too.
A: Since $1/(y+1/z)<1$, while $x,y$ and $z$ are whole numbers, we must make $x$ as big as possible, namely $x=5$. Similarly, the next choice is $y=1$, to minimize $y+1/z$, and then $z=5$. If, in addition, it is required that $z\neq x$, the last choice has to be $z=4$. 
A: $$
f(x,y,z) = \frac{1}{x+\frac{1}{y+\frac{1}{z}}}
$$
is decreasing in $x$ and $z$, and increasing in $y$. Since $x,z$ are distinct integers we have either
$$
 x < z \implies x \le 4, y \ge 1, z \le 5  \implies f(x, y, z) \ge f(4, 1, 5) = \frac{6}{29}
$$
or
$$
 z < x \implies x \le 5, y \ge 1, z \le 4  \implies f(x, y, z) \ge f(5, 1, 4) = \frac{5}{29}
$$
so that the minimal value is the smaller of these two, which is $f(5, 1, 4) = \frac{5}{29}$.
A: It's just
$$\frac{1}{5+\frac{1}{1+\frac{1}{4}}}=\frac{5}{29}.$$
Indeed, we need to prove that
$$\frac{1}{x+\frac{1}{y+\frac{1}{z}}}\geq\frac{5}{29}$$ or
$$\frac{1}{x+\frac{z}{yz+1}}\geq\frac{5}{29}$$ or
$$\frac{yz+1}{xyz+x+z}\geq\frac{5}{29}$$ or
$$5yz(5-x)+5(5-x)+4z(y-1)+4-z\geq0.$$
Now, if $z<4$ so $$5yz(5-x)+5(5-x)+4z(y-1)+4-z>0,$$ which gives
$$\frac{1}{x+\frac{1}{y+\frac{1}{z}}}>\frac{5}{29}$$ and we'll get a greater value than $\frac{5}{29}.$
If $z=5$, so $x<5$ and  we obtain $$25y(5-x)+5(5-x)+20(y-1)+4-5>0,$$ which gives again:
$$\frac{1}{x+\frac{1}{y+\frac{1}{z}}}>\frac{5}{29}.$$
For $z=4$ we obtain $$20y(5-x)+5(5-x)+16(y-1)\geq0,$$
where the equality occurs for $x=5$ and $y=1$ and we are done!
