Does Symmetricity Always Hold?

Write the given expression as $\dfrac{4}{x}+\dfrac{9}{y}+\dfrac{16}{z}$ .

Using Titu's Lemma/Cauchy Schwarz in Engel form- We get- Min(expression)= $\dfrac{(2+3+4)^{2}}{x+y+z}$ . We know that $x+y+z=3$ thus answer = $\dfrac{81}{3}=27$ . I think this is the easiest possible way. Kindly upvote if you liked this :)

For those who dont know Titu Lemma it is nothing but- $\displaystyle \sum_{i=1}^n \dfrac{a_i^2}{b_i} =\dfrac{(a_{1}^{2})}{b_{1}}+...........+\dfrac{(a_{n}^{2})}{b_{n}} \geq \dfrac{(a_{1}+............+a_{n})^{2}}{b_{1}+...........+b_{n}}$

Or, the Cauchy Schwarz Engel form, which is , for sets $(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$ , we have $\displaystyle \left| \sum a_ib_i \right| ^2 \leq \biggl( \sum a_i^2 \biggr) \biggl( \sum b_i^2 \biggr)$

$\displaystyle\E=\frac { 4yz+9xz+16xy }{ xyz } =\frac { 4 }{ x } +\frac { 9 }{ y } +\frac { 16 }{ z } \\ \\ Now\quad using\quad Famous\quad inequality\quad For\quad positive\quad variables\quad \\ Method-1:\quad AM\quad \ge \quad HM\\ \Longrightarrow \frac { \frac { x }{ 2 } +\frac { x }{ 2 } +\frac { y }{ 3 } +\frac { y }{ 3 } +\frac { y }{ 3 } +\frac { z }{ 4 } +\frac { z }{ 4 } +\frac { z }{ 4 } +\frac { z }{ 4 } }{ 9 } \quad \ge \quad \frac { 9 }{ \frac { 2 }{ x } +\frac { 2 }{ x } +\frac { 3 }{ y } +\frac { 3 }{ y } +\frac { 3 }{ y } +\frac { 4 }{ y } +\frac { 4 }{ z } +\frac { 4 }{ z } +\frac { 4 }{ z } } \\ \\ \Longrightarrow x+y+z\quad \ge \quad \frac { 81 }{ E } \\ \Longrightarrow E\quad \ge \quad 27\\ { \Longrightarrow E }_{ min }=27\\ \quad \\ Method-2:\\ Using\quad AM\quad \ge \quad GM\\ \quad \Longrightarrow E\times (3)=E\times (x+y+z)=\quad (4+9+16)\quad +\quad (\frac { 4y }{ x } +\frac { 9x }{ y } )+\quad (\frac { 9z }{ y } +\frac { 16y }{ z } )+\quad (\frac { 4z }{ x } +\frac { 16x }{ z } )\quad \ge \quad 81\\ \Longrightarrow E\quad \ge \quad 27$ .

Does anyone has other method...??

$E = \frac{2^{2}}{x} + \frac{3^{2}}{y} + \frac{4^{2}}{z}$

$E = ( x + y + z)( \frac{2^{2}}{\sqrt{3}x} + \frac{3^{2}}{\sqrt{3}y} + \frac{4^{2}}{\sqrt{3}z}$

Now,

$(\sum a_{i} b{i})^{2} \leq (\sum a_{i})^{2}(\sum b_{i})^{2}$

least when it is equal to $( \frac{2 +3 + 4}{\sqrt{3}})^{2}$

$= \frac{9^{2}}{3} = 27$

$E = \frac {4yz+9xz+16xy}{xyz},$ can be simplified in the form: $\begin{cases}E = \frac 4 x + \frac 9 y + \frac {16} z = 0\\g(x)=x+y+z-3=0\end{cases}\,(1)$ that can be minimized with Lagrange's multipliers method: $F=E-{\lambda}g(x),\,\lambda \in \mathbb{R},$ $\begin{cases}\frac {\partial F}{\partial x} = 0\\\frac {\partial F}{\partial y} = 0 \\\frac {\partial F}{\partial x} = 0\\\frac {\partial F}{\partial {\lambda}} = 0\end{cases}\,(2)$ which leads to: $\begin{cases}{\lambda}x^2+4=0\\{\lambda}y^2+9=0\\{\lambda}z^2+16=0\\x+y+z-3=0\end{cases}\,(3)$ with solutions: $\begin{cases}\lambda =-9;\,x=\frac 2 3;\,y=1;\,z=\frac 4 3;\\\lambda =-\frac {25} 9;\,x=-\frac 6 5;\,y=\frac 9 5;\,z=\frac {12} 5\\\lambda =-1;\,x=2;\,y=-3;\,z=4;\\\lambda =-\frac 1 9;\,x=6;\,y=9;\,z=-12;\end{cases}$ only the first solution with $x,\,y,\,z,\,$ positives is acceptable and leads to: $E_{min}=\boxed {27}$