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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.

nar
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4 Answers4

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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.

Mark Bennet
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  • Yes, make sense, but I hope I can get why those numbers is the answer, I thought in the same way, cause this problem is for grade 11 and it is an Olympiad problem, I think maybe there is an algebraic method to proof this. – nar Aug 10 '19 at 11:49
  • @nar Any proof will do - it doesn't need to be complicated. Olympiad proofs are often simple once you find them. It is the finding which is tricky. – Mark Bennet Aug 10 '19 at 12:31
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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 $z=4$.

John Bentin
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$$ f(x,y,z) = \frac{1}{x+\frac{1}{y+\frac{1}{z}}} $$ 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 $f(5, 1, 4) = \frac{5}{29}$.

Martin R
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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!

Michael Rozenberg
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