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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.

Given positive variables $x$, $y$ and $z$ and positive parameters $a, b, c$.

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

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

Well...

Perhaps delete what I had said 288 days ago. Let's start this all over again.

I can't leave this question to go waste. I shouldn't have overgeneralised this inequality.

Here is the correct inequality.

Given positive $x, y, z$ and distinct parameters $m, n, p > 0$. Calculate the minimum value of $$\large \left(\frac{n + p}{mx} + \frac{p + m}{ny} + \frac{m + n}{pz}\right) \cdot \sqrt{yz + zx + xy}$$

I have provided a solution below and I would be greatly appreciated if anyone could come up with a better solution than mine.

I apologise for the misunderstanding.

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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 '19 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 '19 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 '19 at 10:33
  • $\begingroup$ “It was simple, yet difficult.” – What is that supposed to mean? $\endgroup$ – Martin R Mar 23 '19 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 '19 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 '19 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 '19 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 '19 at 7:20
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    $\begingroup$ What is the difference between wrong and totally wrong? $\endgroup$ – Dr. Sonnhard Graubner Mar 23 '19 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 '19 at 7:33
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Let $a = mx, b = ny, c = pz$. We have that $\dfrac{n + p}{mx} + \dfrac{p + m}{ny} + \dfrac{m + n}{pz}$

$$ = \frac{n + p}{a} + \frac{p + m}{b} + \frac{m + n}{c} = \left(\frac{1}{b} + \frac{1}{c}\right) \cdot m + \left(\frac{1}{c} + \frac{1}{a}\right) \cdot n + \left(\frac{1}{a} + \frac{1}{b}\right) \cdot p$$

$$ = \frac{b + c}{bc} \cdot m + \frac{c + a}{ca} \cdot n + \frac{a + b}{ab} \cdot p$$

$$ \implies \left(\dfrac{n + p}{mx} + \dfrac{p + m}{ny} + \dfrac{m + n}{pz}\right) \cdot \sqrt{yz + zx + xy}$$

$$ = \left(\frac{b + c}{bc} \cdot m + \frac{c + a}{ca} \cdot n + \frac{a + b}{ab} \cdot p\right) \cdot \sqrt{\frac{bc}{np} + \frac{ca}{pm} + \frac{ab}{mn}}$$

$$ \ge \sqrt{3\left(\frac{bc}{np} + \frac{ca}{pm} + \frac{ab}{mn}\right)\left[\frac{np(c + a)(a + b)}{bca^2} + \frac{pm(a + b)(b + c)}{cab^2} + \frac{mn(b + c)(c + a)}{abc^2}\right]}$$

$$ \ge \sqrt{3} \cdot \left[\frac{\sqrt{(c + a)(a + b)}}{a} + \frac{\sqrt{(a + b)(b + c)}}{b} + \frac{\sqrt{(b + c)(c + a)}}{c}\right]$$

$$ \ge \sqrt{3} \cdot \left(\frac{\sqrt b + \sqrt c}{\sqrt a} + \frac{\sqrt c + \sqrt a}{\sqrt b} + \frac{\sqrt a + \sqrt b}{\sqrt c}\right) \ge 3\sqrt 3$$

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