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. 

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!
