Minima doesn't always involve symmetry!

This is Similar question which I posted . Click here

So Here also I post same Solution

Let $E=\frac { 16yz+36xz+64xy }{ xyz } =\frac { 16 }{ x } +\frac { 36 }{ y } +\frac { 64 }{ z }$ .

Now using $AM\quad \ge \quad HM$ .

$\dfrac { \frac { x }{ 4 } +\frac { x }{ 4 } +\frac { x }{ 4 } +\frac { x }{ 4 } +\frac { y }{ 6 } +\frac { y }{ 6 } +\frac { y }{ 6 } +\frac { y }{ 6 } +\frac { y }{ 6 } +\frac { y }{ 6 } +\frac { z }{ 8 } +\frac { z }{ 8 } +\frac { z }{ 8 } +\frac { z }{ 8 } +\frac { z }{ 8 } +\frac { z }{ 8 } +\frac { z }{ 8 } +\frac { z }{ 8 } }{ 18 } \\ \quad \\ \ge \quad \frac { 18 }{ \frac { 4 }{ x } +\frac { 4 }{ x } +\frac { 4 }{ x } +\frac { 4 }{ x } +\frac { 6 }{ y } +\frac { 6 }{ y } +\frac { 6 }{ y } +\frac { 6 }{ y } +\frac { 6 }{ y } +\frac { 6 }{ y } +\frac { 8 }{ z } +\frac { 8 }{ z } +\frac { 8 }{ z } +\frac { 8 }{ z } +\frac { 8 }{ z } +\frac { 8 }{ z } +\frac { 8 }{ z } +\frac { 8 }{ z } + } \\ \\ \Longrightarrow x+y+z\quad \ge \quad \frac { 324 }{ E } \\ \Longrightarrow E\quad \ge \quad 54\\ { \Longrightarrow E }_{ min }=54$ .

Well this Can also be solved by using AM-GM inequality

$Using\quad AM\quad \ge \quad GM\\ E\times (6)=E\times (x+y+z)\\ \quad \quad \quad \quad =\quad (16+36+64)\quad +\quad (\frac { 16y }{ x } +\frac { 36x }{ y } )+\quad (\frac { 36z }{ y } +\frac { 64y }{ z } )+\quad (\frac { 16z }{ x } +\frac { 64x }{ z } )\quad \ge \quad 324$ .

Now Use simple AM-GM on variables and get The required answer.

NOTE

Values of $x,y,z$ . are $( 4/3 , 2 , 8/3 )$ .

( By using fact that numbers are equal when AM=GM=HM ).

This question is a direct application of Titu's Lemma.

Simplifying the expression, we get:

$\frac{16}{x}+\frac{36}{y}+\frac{64}{z}$

Applying Titu's Lemma we get the minimum value of the given expression as:

$\frac{(\sqrt{16}+\sqrt{36}+\sqrt{64})^{2}}{x+y+z}$

$\frac{(4+6+8)^{2}}{6}$

$\frac{324}{6}=54$

and we are done!