Calculate the minimum value of $\left(\frac{n + p}{mx} + \frac{p + m}{ny} + \frac{m + n}{pz}\right) \cdot \sqrt{yz + zx + xy}$.

Question

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.

Answer 1

Let $a = mx, b = ny, c = pz$. We have that $\dfrac{n + p}{mx} + \dfrac{p + m}{ny} + \dfrac{m + n}{pz}$

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

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

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

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

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

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

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

Answer 2

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$$\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)\sqrt{xy+yz+zx}\geq 3\sqrt{3}\sqrt [3]{abc}$$