# $10(x + y) + 4z = 5xyz$. Find the min value of: $A = 2x + y + 2z$

PROBLEM: We have $x,y,z \in \mathbb{R}^+$ such as $10(x + y) + 4z = 5xyz$. Find the min value of: $$A = 2x + y + 2z$$

My attempt:

I would find the min of: $$A^2 = \frac{(2x+y+2z)^2(10x+10y+4z)}{5xyz}$$ and then let $z=1$, after that I used Lagrange method to find min of: $$\frac{(2x+y+2)^2(10x+10y+4)}{5xy}$$

Finally I found the result which is min value of $A$ when $\dfrac{x}{4}=\dfrac {y}{6}=\dfrac{z}{5}$

But the problem is, my brother is just a high school student so he can't understand Lagrange's method, so I wonder if having another simple method for this?

• I think something is wrong in your inequality occuring. – Michael Rozenberg Jun 15 '17 at 10:21
• there is mistake somewhere in your solution. If $z=1,$ then $x = 1$ and $y = 1.2$, but that does not satisfy the given condition. – dezdichado Jun 15 '17 at 10:31
• no, it's just a ratio among them, the exact result is $(x,y,z)=\frac{1}{\sqrt{5}}(4,6,5)$ – Alex N Jun 15 '17 at 10:39
• @MichaelRozenberg yeah, sorry, it is $4:6:5$ – Alex N Jun 15 '17 at 10:40

For $x=\frac{4}{\sqrt5}$, $b=\frac{6}{\sqrt5}$ and $c=\sqrt5$, we get a value $\frac{24}{\sqrt5}$.
Indeed, let $x=\frac{4a}{\sqrt5}$, $y=\frac{6b}{\sqrt5}$ and $z=\sqrt5c$.
Hence, the condition gives $2a+3b+c=6abc$ and we need to prove that $$4a+3b+5c\geq12$$ or $$\frac{(4a+3b+5c)^2(2a+3b+c)}{6abc}\geq144$$ or $$(4a+3b+5c)^2(2a+3b+c)\geq864abc,$$ which is true by AM-GM: $$(4a+3b+5c)^2(2a+3b+c)\geq\left(12\sqrt[12]{a^4b^3c^5}\right)^2\cdot6\sqrt[6]{a^2b^3c}=864abc.$$ The equality occurs for $a=b=c=1$.