A Hard Inequality

Question

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)

Answer 1

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!

Answer 2

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 $L ={15\over 2}$.