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 $\lambda$ ? 
 A: Let $z=\sqrt{xy}t$.
Thus, by AM-GM
$$\frac{(x+y+z)((x+y)^2+(x+y+4z)^2)}{xyz}\geq\frac{(2\sqrt{xy}+z)(4xy+(2\sqrt{xy}+4z)^2)}{xyz}=$$
$$=\frac{(2+t)(4+(2+4t)^2)}{t}=100+\left(\frac{8(t+2)(2t^2+2t+1)}{t}-100\right)=$$
$$=100+\frac{4(2t-1)^2(t+4)}{t}\geq100.$$
The equality occurs for $$x=y=2z,$$ which says that we got a minimal value of the first expression, which says that a maximal value of $\lambda,$ for which our inequality is true for any positives $x$, $y$ and $z$ it's $100.$
A: 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 $9$!!!
$$\sqrt[9]{259200}\times 9\times 2>40$$!!!
A: We seek to minimize $$f(x,y,z)=\frac{x+y+z}{xyz}\left((x+y)^2+(x+y+4z)^2\right)$$  If $x\neq y$ replace both $x$ and $y$ by ${x+y\over2}$  Observe that the numerator doesn't change, but the denominator increases, since $$xy\leq\left({x+y\over2}\right)^2.$$  Therefore, the value of $f$ decreases, and we can assume from now on that $x=y$. 
This should make it easier to deal with.
Now if we write $z=tx$, the $x$'s cancel in numerator and denominator, and we are left with a function of $t$ alone.  Carrying out this program we have $$\begin{align}
h(t)=g(x,tx)&={2x+tx\over tx^3}\left(4x^2+(2+4t)^2x^2\right)\\
&={2+t\over t}\left(4+(2+4t)^2\right)\\
&={(2+t)\left(16t^2+16t+8\right)\over t}\\
&={16t^3+48t^2+40t+16\over t}\\
&=16t^2+48t+40+{16\over t}
\end{align}$$
so that $$h'(t)=32t+48-{16\over t^2}$$ and $$h'(t)=0\implies 2t^3+3t^2-1=0\implies (t+1)^2(2t-1)=0$$
so that the the only positive solution is $t=\frac12,$ and we find that the minimum is $$h\left(\frac12\right)={16\over4}+{48\over2}+40+2\cdot16=100$$ achieved when $$2x=2y=z.$$
