<s>[INST] Solving an inequality with three variables Find the largest integer $\lambda$ such that :
$$\frac{\lambda (xyz)}{x+y+z} \le(x+y)^2 + (x+y+4z)^2.$$
Here $x,y$ and $z$ are positive real numbers
The answer is only achieved using calculus. Can anyone use AM-GM and get $\lambda$ ? 
 [/INST] I think it is $$\frac{x+y+z}{xyz}\left((x+y)^2+(x+y+4z)^2\right)\geq 100$$ and the equal sign holds if $$x=y=1,z=\frac{1}{2}$$
This is true since we have $$\frac{x^3+3x^2y+3xy^2+y^3+5x^2z+5y^2z+12xz^2+12yz^2+8z^3}{9}\geq \sqrt[9]{259200x^9y^9z^9}$$
Now it must
$$9\times\sqrt[9]{259200}\times 2\geq 40$$
I forgot the factor $9$!!!
$$\sqrt[9]{259200}\times 9\times 2>40$$!!!
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