Find the largest constant $k$ such that $\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$ 
Find the largest constant $k$ such that $$\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$$

My attempt, 
By A.M-G.M, $$(a+b)^2+(a+b+4c)^2=(a+b)^2+(a+2c+b+2c)^2$$
$$\geq (2\sqrt{ab})^2+(2\sqrt{2ac}+2\sqrt{2bc})^2$$
$$=4ab+8ac+8bc+16c\sqrt{ab}$$
$$\frac{(a+b)^2+(a+b+4c)^2}{abc}\cdot (a+b+c)\geq\frac{4ab+8ac+8bc+16c\sqrt{ab}}{abc}\cdot (a+b+c)$$
$$=(\frac{4}{c}+\frac{8}{b}+\frac{8}{a}+\frac{16}{\sqrt{ab}})(a+b+c)$$
$$=8(\frac{1}{2c}+\frac{1}{b}+\frac{1}{a}+\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ab}})(\frac{a}{2}+\frac{a}{2}+\frac{b}{2}+\frac{b}{2}+c)$$
$$\geq 8(5\sqrt[5]{\frac{1}{2a^2b^2c}})(5\sqrt[5]{\frac{a^2b^2c}{2^4}})=100$$
Hence the largest constant $k$ is $100$
Is my answer correct? Is there another way to solve it? Thanks in advance.

One more way...  
Noting that replacing $a, b$ with $\frac{a+b}2, \frac{a+b}2$ leaves RHS unchanged but increases the LHS, we have to only check for the case $a=b$.  Further as the inequality is homogeneous in $a, b, c$; WLOG we may set $a=1$.  Hence we need only look for the minimum of the univariate 
$$f(c) = (4+(2+4c)^2)\cdot \frac{2+c}c$$
This is easily done using calculus, or by noting $f(c) = \dfrac{4(4+c)(2c-1)^2}c + 100 \geqslant 100$.
