# Find the minimum value of $\left(\frac{a}{x} + \frac{b}{y} + \frac{c}{z}\right)\sqrt{yz + zx + xy}$

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.

• In the general case the answer is very ugly and it's just impossible to write it. – Michael Rozenberg Mar 23 at 7:18
• By the way, for $(a,b,c)=(1,2,5)$ we can get a nice answer. – Michael Rozenberg Mar 23 at 7:37
• Now that's what they asked in the competition for the participants in the lower grade that same year. – Lê Thành Đạt Mar 23 at 10:33
• “It was simple, yet difficult.” – What is that supposed to mean? – Martin R Mar 23 at 10:52
• There are only 2 lines to ask for the problem but nobody can solve it. – Lê Thành Đạt Mar 23 at 10:53

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$$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\sqrt{xy+yz+zx}\geq 3\sqrt{3}\sqrt [3]{abc}$$
• 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. – Macavity Mar 23 at 6:48
• @Sonnhard Your reasoning is total wrong. Try to understand when does the equality occur for different $a$, $b$ and $c$. – Michael Rozenberg Mar 23 at 7:20