Geometrical Inequalities!

@Calvin Lin , Sir here it is.

I will prove

$\large\ x(1 - y^2)(1 - z^2) + y(1 - z^2)(1 - x^2) + z(1 - x^2)(1 - y^2) = 4xyz$

subject to $x + y + z = xyz$ .

The conclusion is immediate if $xyz = 0,$ so we may assume that $x, y, z \neq 0$ .

Dividing through by $4xyz$ we transform the desired equality into

$\large\ \frac { 1 - { y }^{ 2 } }{ 2y } \cdot \frac { 1 - { z }^{ 2 } }{ 2z } + \frac { 1 - { z }^{ 2 } }{ 2x } \cdot \frac { 1 - { x }^{ 2 } }{ 2x } + \frac { 1 - { x }^{ 2 } }{ 2x } \cdot \frac { 1 - { y }^{ 2 } }{ 2y } = 1$ .

This, along with the condition from the statement, makes us think about the substitutions $x = \tan A, y = \tan B, z = \tan C$ , where $A, B, C$ are the angles of a triangle. Using the double-angle formula

$\large\ \frac { 1 - \tan ^{ 2 }{ u } }{ 2\tan { u } } = \cot { u }$ ,

we further transform the equality into

$\large\ \cot {2B} \cot {2C} + \cot {2C} \cot {2A} + \cot {2A} \cot {2B} = 1.$

But this is equivalent to

$\large\ \tan {2A} + \tan {2B} + \tan {2C} = \tan {2A} \tan {2B} \tan {2C},$

which follows from $\large\ { \tan{(2A + 2B + 2C)} = \tan {2π} = \boxed 0 }$ .

So, i just replaced $4xyz$ by $kxyz$ and asked for $k$ .