A Hard Inequality Given that $x,y,z$ are positive real numbers such that $2x+4y+7z=2xyz$, find the minimum of $L=x+y+z$.
Does anybody have a solution that is purely algebraic?
I was only able to solve it with Lagrange multipliers.
Also, how would you show that the solution given by Lagrange multipliers is in fact a global solution?
Note: By a change of variables, this is equivalent to minimizing $$L=a+b+c-\frac{3}{2}$$ subject to $$2 a b c = a + 4 b + 2 a b + 7 c + a c - 9$$
where $a>0,b>\frac{1}{2},c>1$.
$L$ is minimized when $a=b=c=3$ and $L=7.5$.
Source: https://brilliant.org/problems/another-weird-inequality/
(I did not write this question)

For $x=3$, $y=2.5$ and $z=2$ we get the value $7.5$.
We'll prove that it's a minimal value.
Indeed, let $x=3a$, $y=2.5b$ and $z=2c$.
Thus, the condition gives 
$$3a+5b+7c=15abc$$ and we need to prove that
$$6a+5b+4c\geq15$$ or
$$(6a+5b+4c)^2(3a+5b+7c)\geq15^3abc,$$ which is true by AM-GM:
$$(6a+5b+4c)^2(3a+5b+7c)\geq\left(15\sqrt[15]{a^6b^5c^4}\right)^2\cdot15\sqrt[15]{a^3b^5c^7}=15^3abc.$$
Done!
