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)
 A: 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!
A: Express $x$: $$x =
{4y+7z\over 2(yz-1)}$$ We get rational function in $y$ with parameter $z$: $$ L = {2y^2z+2yz^2+2y+5z\over
2(yz-1)}$$
We are searching for local minimum $m$ of $L$ (which is greater then it local maximum). So the equation $L=m$ must have exactly one solution (at fixed $z$) on $y$ thus the discriminat $D_y$ of:
$$ 2y^2z+2y(z^2-mz+1)+5z+2m=0$$
must be $0$, so (I skip some calculation here) 
$$m^2z^2-2mz(z^2+3)+(z^4-8z^2+1)=0$$
If we expres $m$ we get (we choose positive sign (obviously?)):
$$m = {z^2+3+ \sqrt{(14z^2+8)}\over z}$$
Here I could not skip the derivate (actually I could, but it is bothersome). If you calculate it you see we get $z=2$ ... and $L ={15\over 2}$.
