# Proving $30x + 3y^2 + \frac{2z^3}{9} + 36 (\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}) \ge 84$ [closed]

If $$x, y, z$$ are positive real numbers, prove that $$30x + 3y^2 + \frac{2z^3}{9} + 36 \left(\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}\right) \ge 84.$$

I genuinely have no clue on how to proceed. Is it proved using repeated CS?

• You're looking for a solution $(x,y,z)$ I assume? Presumably with a least one variable free? And given teh number theory tage, presumably you want integer solutions? Aug 22, 2022 at 0:51
• Uh, no. This is proving an inequality with 3 unknown variables. Aug 22, 2022 at 0:54
• I guess, you can interpret the problem as constrained minimization problem in the domain of $\mathbb{R}^3_{>0}$. After that, finding the local minimum, you’d probably come to it being $84$. Aug 22, 2022 at 1:04
• Oh... you want to solve the entire equality? Huh. Um. Yeah, that's gonna be really messy. Aug 22, 2022 at 1:04
• This needs some context (ref. above comment). It sounds like a really good question. Personally : given it's a contest question , all you will need to do is mention the textbook/PDF etc. where you found this, and one reference that you've been using to read up three-variable inequalities. If I may suggest one : I think Zdravko Cvetkovski's "Inequalities" contains some identities that may be useful for your purpose. Aug 22, 2022 at 6:45

Remark: Once we know the equality case $$x = 1, y = 2, z = 3$$, we apply AM-GM.

Using AM-GM, we have \begin{align*} &30\cdot x + 12 \cdot (y/2)^2 + 6\cdot (z/3)^3 + 18 \cdot \frac{1}{xy/2} + 6\cdot \frac{1}{yz/6} + 12 \cdot \frac{1}{zx/3}\\ \ge\,& 84\sqrt{x^{30}\cdot (y/2)^{24}\cdot (z/3)^{18} \cdot \left(\frac{1}{xy/2}\right)^{18}\left(\frac{1}{yz/6}\right)^6\left(\frac{1}{zx/3}\right)^{12} }\\ =\,& 84. \end{align*}

We are done.

• Not OP, but can you elaborate on how you figured out you could rearrange terms in that manner and then apply AM-GM twice after rearranging to get the desired result? It just seems kind of arbitrary, because the $-yz + 2/9 z^3 + 3y^2 - 84$ is a little too awkward to ignore, yet you proceed with AM-GM regardless. So I am just looking for any motivation or patterns you can share. Aug 22, 2022 at 7:12
• @politeproofs I added some explanation. Aug 22, 2022 at 7:20
• @politeproofs I just found it is just AM-GM directly. Aug 22, 2022 at 8:30

If we want to find the minimum, I think we can create a system from the implicit derivatives. We can multiply through by $$xyz$$, then subtract the right side to get:

$$f(x,y,z) = 30x^2yz+3xy^3z+\textstyle{\frac29}xyz^4+36x+36y+36z-84xyz \geqslant 0$$

And we want to show that this function has a minimum of $$0$$. Then we have:

\begin{align} \frac{\partial f}{\partial x} &= 60xyz+3y^3z+ \textstyle{\frac{2}{9}}yz^4-84yz+36 \\ \frac{\partial f}{\partial y} &= 30x^2z+9xy^2z+ \textstyle{\frac{2}{9}}xz^4 -84xz+36\\ \frac{\partial f}{\partial z} &= 30x^2y+3xy^3+ \textstyle{\frac{8}{9}}xyz^3-84xy+36 \end{align}

To find minima, we want all of these partial derivatives to be equal to $$0$$. This is a system of three equations in three variables:

\left\{ \begin{aligned} 60xyz+3y^3z+ \textstyle{\frac{2}{9}}yz^4-84yz &= -36 \\ 30x^2z+9xy^2z+ \textstyle{\frac{2}{9}}xz^4 -84xz &= -36 \\ 30x^2y+3xy^3+ \textstyle{\frac{8}{9}}xyz^3-84xy &= -36 \end{aligned} \right.

That's a bit... ugly. Let's try something:

\left\{ \begin{aligned} 60x^2yz+3xy^3z+ \textstyle{\frac{2}{9}}xyz^4-84xyz +36x &= 0 \\ 30x^2yz+9xy^3z+ \textstyle{\frac{2}{9}}xyz^4-84xyz +36y &= 0 \\ 30x^2yz+3xy^3z+ \textstyle{\frac{8}{9}}xyz^4-84xyz +36z &= 0 \end{aligned} \right.

Now set $$xyz=a$$ and we have:

\left\{ \begin{aligned} 60ax+3ay^2z+ \textstyle{\frac{2}{9}}az^3 +36x &= 84a \\ 30ax+9ay^2z+ \textstyle{\frac{2}{9}}az^3 +36y &= 84a \\ 30ax+3ay^2z+ \textstyle{\frac{8}{9}}az^3 +36z &= 84a \end{aligned} \right.

I do believe we'll need to solve the system by substitution. I'll leave this here for the moment and return.