MHB Proving $(x^2+1)(y^2+4)(z^2+9) \ge 100$ for $6x+3y+2z=10+xyz$

[anemone]

Let $x,\,y,\,z$ be real numbers such that $6x+3y+2z=10+xyz$.

Prove that $(x^2+1)(y^2+4)(z^2+9)\ge 100$.

 

[Albert1]

Let $x,\,y,\,z$ be real numbers such that $6x+3y+2z=10+xyz---(1)$.

Prove that $k=(x^2+1)(y^2+4)(z^2+9)\ge 100$.

Spoiler

$(x,y,z)=(1,1,1)(0,0,5)$,satisfy $(1)$
when $(x,y,z)=1,1,1), k=100$
when $(x,y,z)=(0,0,5),k=136>100$
now we will prove $k<100$ is out of the question
suppose we can find one set of $x,y,z$ satisfy (1) and $k<100$
then we have :
$ 100>k\geq 36(x^2+1),100>k\geq 9(y^2+4),100>k\geq 4(z^2+9)$ (for all $x,y,z)$
that means for each
$\dfrac{-8}{6}\leq x \leq \dfrac {8}{6}$
$\dfrac{-8}{3}\leq y \leq \dfrac {8}{3}$
$-4\leq z \leq 4$ and satisfy (1)
it should be :$ 100\times 36^2>36^2k= 36^2(x^2+1)(y^2+4)(z^2+9)$
now we check point $x=0,y=2,z=2$
$36^2\times100>36^2\times 1\times 8\times 13=36^2\times 104$
this is a contradiction ,so we conclude that $k\geq 100$

 

[anemone]

Thanks Albert for participating in this challenge.

Solution of other:

Spoiler

Note that

$(x^2+1)(y^2+4)(z^2+9)-(6x+3y+2z-xyz)^2$

$=9 x^2 y^2+4 x^2 z^2+y^2 z^2+12 x^2 y z+6 x y^2 z+4 x y z^2-36 x y-24 x z-12 y z+36$

$=(4 x^2 z^2+y^2 z^2+4 x y z^2)+(9 x^2 y^2-36 x y+36)+(12 x^2 y z+6 x y^2 z-24 x z-12 y z)$

$=(2xz+yz)^2+9(xy-2)^2+6(2 x^2 y z+x y^2 z-4 x z-2 y z)$

$=(2xz+yz)^2+6(2xz+yz)(xy-2)+9(xy-2)^2$

$=(2xz+yz+3(xy-2))^2$

And since we're given $6x+3y+2z-xyz=10$, we can conclude that

$(x^2+1)(y^2+4)(z^2+9)=(6x+3y+2z-xyz)^2+(2xz+yz+3(xy-2))^2\ge 100$