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

Question

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?

Answer 1

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[84]{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.

Answer 2

Partial answer:

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.