Find the minimum $k$ 
Find the minimum $k$, which $\exists a,b,c>0$, satisfies
  $$ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$$

My Progress
With the help of Mathematica, I found that when $k=100$, we can take $a=1,b=1,c=1/2$. And I'm pretty sure that $k=100$ is the answer, but I couldn't prove it.
 A: Effectively, you want to show
$$\frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c) \geqslant 100$$
and you already have a case of equality.
Using homogeneity, we may set $a+b+c=5$, to equivalently show
$$(5-c)^2+(5+3c)^2 \geqslant 20 abc$$
Now $a+b = 5-c$, so for any $c$, we have $ab$ maximized when $a=b$.  Thus it is enough to show
$$(5-c)^2+(5+3c)^2 \geqslant 20 \left(\frac{5-c}2\right)^2c$$
which leads us to the polynomial inequality $5(c-1)^2(10-c) \geqslant 0$ which is obvious.
A: Let $a=b=2$ and $c=1$. 
Hence, $k\geq100$.
We'll prove that $100$ it's an answer. 
Indeed, let there are positives $a$, $b$ and $c$ for which $ \frac{kabc}{a+b+c}\geq (a+b)^2+(a+b+4c)^2$ and $k<100$.
But it's impossible because we'll prove now that $ \frac{100abc}{a+b+c}\leq (a+b)^2+(a+b+4c)^2$.
Let $c=(a+b)x$. 
Hence, by AM-GM $\frac{100abc}{a+b+c}\leq\frac{25(a+b)^2c}{a+b+c}=\frac{25(a+b)^2x}{x+1}$.
Thus, it remains to prove that $\frac{25(a+b)^2x}{x+1}\leq(a+b)^2+(a+b+4(a+b)x)^2$ or
$\frac{25x}{x+1}\leq1+(1+4x)^2$, which is AM-GM again or 
$(4x-1)^2(x+2)\geq0$. Done!
