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Going back a few more years and you can find more and more interesting problems over the years as time turns back. I am still surprised at how easy this competition has become. Then I come across this problem, which goes by the following.

$x$, $y$ and $z$ are positive variables and $a = F_{n - 1}$, $b = F_{n + 1}$ are positive parameters. ($F_n$ is the $n^{th}$ Fibonacci number.

Find the minimum value of $\left(\dfrac{1}{x} + \dfrac{a}{y} + \dfrac{b}{z}\right)\sqrt{yz + zx + xy}$.

It was simple, yet difficult. I wished to find a solution without a solution without using Lagrange multipliers but found no results. I would be grateful if you have a solution like so.

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    $\begingroup$ In the general case the answer is very ugly and it's just impossible to write it. $\endgroup$ – Michael Rozenberg Mar 23 at 7:18
  • $\begingroup$ By the way, for $(a,b,c)=(1,2,5)$ we can get a nice answer. $\endgroup$ – Michael Rozenberg Mar 23 at 7:37
  • $\begingroup$ Now that's what they asked in the competition for the participants in the lower grade that same year. $\endgroup$ – Lê Thành Đạt Mar 23 at 10:33
  • $\begingroup$ “It was simple, yet difficult.” – What is that supposed to mean? $\endgroup$ – Martin R Mar 23 at 10:52
  • $\begingroup$ There are only 2 lines to ask for the problem but nobody can solve it. $\endgroup$ – Lê Thành Đạt Mar 23 at 10:53
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Hint: $$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\geq 3\sqrt[3]\frac{abc}{xyz}$$ and $$yz+zx+xy\geq 3\sqrt[3]{(xyz)^2}$$ Putting things together we obtain$$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\sqrt{xy+yz+zx}\geq 3\sqrt{3}\sqrt [3]{abc}$$

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    $\begingroup$ The equal sign doesn't occur like so. $\endgroup$ – Lê Thành Đạt Mar 23 at 5:57
  • $\begingroup$ For equality this needs $a=b=c$, which is not assured as these are given parameters, hence this does not give the minimum, only a lower bound. $\endgroup$ – Macavity Mar 23 at 6:48
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    $\begingroup$ @Sonnhard Your reasoning is total wrong. Try to understand when does the equality occur for different $a$, $b$ and $c$. $\endgroup$ – Michael Rozenberg Mar 23 at 7:20
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    $\begingroup$ What is the difference between wrong and totally wrong? $\endgroup$ – Dr. Sonnhard Graubner Mar 23 at 7:27
  • $\begingroup$ It seems you have right, this is only a lower bound, when i use calculus, the solution is pages long. $\endgroup$ – Dr. Sonnhard Graubner Mar 23 at 7:33

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