Asymmetric inequality

$\large \text{Key Technique: Examine the equality case.}$

The main idea here is the asymmetric factor . It is considerably motivated to break up the term $4c$ among $a$ and $b$ somehow to make stuff a bit more "even out" so to speak. So we suspect that equality is when $a=b=2c$ which gives us $k=100$ .

This equality case resembles that of AM-GM . Part of our strategy is to break up $4c$ as in the equality case. So we proceed as follows.

$\frac{a+b+c}{abc} \times [(a+b)^2 + (a+b+4c)^2]$

$= \frac{a+b+c}{abc} \times [(a+b)^2 + (a+2c+b+2c)^2]$

$\geq \frac{a+b+c}{abc} \times [4ab + (2\sqrt{2ac} + 2\sqrt{2bc})^2]$

$= \frac{a+b+c}{abc} \times [4ab + 8ac + 8bc + 16 \sqrt{ac} \times c ]$

$= (a+b+c) \left( \frac{4}{c} + \frac{8}{a} + \frac{8}{b} + \frac{16}{\sqrt{ab}} \right)$

Up till now we will notice that the variables on the denominator do not immediately cancel out if we just whack AM-GM straight . One way to remedy this is to break up the terms further . Since the numerator involve powers of $2$ , we break stuff up into halves . We are also motivated to break things up into $5$ parts to create the factors of $5$ in the equality case when $k=100$ .

$= \left( \frac{a}{2} + \frac{a}{2} + \frac{b}{2} + \frac{b}{2} + c \right) \left( \frac{4}{c} + \frac{8}{a} + \frac{8}{b} + \frac{8}{\sqrt{ab}} + \frac{8}{\sqrt{ab}}\right)$

$\geq \left( 5 \sqrt[5]{\frac{a^2b^2c}{2^{4}}} \right) \times \left( 5 \sqrt[5]{\frac{2^{14}}{a^2b^2c}} \right)$

$= 100$

if $a,b,c>0$ ,I have answer $k\le 100$

since Use AM-GM inequality we have $\begin{aligned}(a+b)^2+(a+b+4c)^2&=(a+b)^2+[(a+2c)+(b+2c)]^2\ge (2\sqrt{ab})^2+(2\sqrt{2ac}+2\sqrt{2bc})^2\\ &=4ab+8ac+8bc+16c\sqrt{ab} \end{aligned}$ so $\dfrac{(a+b)^2+(a+b+4c)^2}{abc}\ge8\left(\dfrac{1}{2c}+\dfrac{1}{b}+\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}+\dfrac{1}{\sqrt{ab}}\right)$ $a+b+c=\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b}{2}+\dfrac{b}{2}+c$ so use AM-GM inequality we have $\dfrac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c)\ge 8\left(5\sqrt[5]{\dfrac{1}{2a^2b^2c}}\right) \left(5\sqrt[5]{\dfrac{a^2b^2c}{2^4}}\right)=100$ if and only if $a=b=2c$

Another solution:
We can assume that: $a+b+c\quad =\quad 1$ (WLOG)
And use AM-GM: $ab\le \frac { { (a+b) }^{ 2 } }{ 4 } =\frac { { (1-c) }^{ 2 } }{ 4 }$
Hence: $k\le 4\frac { { (1-c) }^{ 2 }+{ (1+3c) }^{ 2 } }{ c{ (1-c) }^{ 2 } } \quad =\quad f(c)$ with (0<c<1)
Not too difficult, we have min f(c) = 100 at c = 0.2
So max k = 100.

An easy way to solve the problem is first rearrange a bit, K<= a+b+c/abc *(a+b)^2 +(a+b+4c)^2 Now, as we can see that in rhs a+b+c/abc is increases if a,b,c all are less than one and the other term has least value equal to 0 but in that case any of one guy must be zero which is not possible so the minima of the rhs will occur when a=b as by symmetry a,b hold same position then pit c=xa and find the value of x so that rhs is minimum , u will get x= 1/2..