Inequality $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ for $a,b,c \in\mathbb{R}$ Find biggest constans k such that  $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ is true for any $a,b,c \in\mathbb{R}$
Could you check up my solution? I'm not sure it's ok -
$(a+b)^2 + (a+b+4c)^2 \ge 0$
and 
$0\ge \frac{kabc}{a+b+c}$
so
$0 \ge k$
 A: We need to worry only about $a+b+c> 0^\dagger$ and the inequality is homogeneous, so we may set $a+b+c=1$ and $c \in (0, 1]$  
Then we have $ab \le \frac14(1-c)^2$ and we need to find largest $k$ satisfying 
$$(1-c)^2+(1+3c)^2 \ge \frac{k}4 c(1-c)^2 \implies \frac{k}4 \le \frac1c+\frac1c\left(\frac{1+3c}{1-c}\right)^2$$
Minimising that gets you $k \le 100$ and when $a=b=\frac25, c=\frac15$, we have equality, so $k=100$ is indeed the largest possible.

$^\dagger$ if $a, b, c \in \mathbb R$, cases with $a+b+c=0$ or $\dfrac{abc}{a+b+c} \to \infty$ occur and no $k$ will satisfy the inequality.
A: Using $\text{arithmetic mean} \geq \text{geometric mean}$, we get
$$
\begin{eqnarray}
(a+b)^2+(a+b+4c)^2&=&(a+b)^2+(a+2c+b+2c)^2 \\
&\geq& \big(2\sqrt{ab}\big)^2+\big(2\sqrt{2ac}+2\sqrt{2b}\big)^2 \\
&=& 4ab+8ac+8bc+16c\sqrt{ab}
\end{eqnarray}
$$
Therefore
$$
\begin{eqnarray}
\frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c)&\geq&\frac{4ab+8ac+8bc+16c\sqrt{ab}}{abc}(a+b+c)\\
&=&\Bigg(\frac4c+\frac8b+\frac8a+\frac{16}{\sqrt{ab}}\Bigg)(a+b+c)\Bigg(\frac a 2+\frac a 2+\frac b 2+\frac b 2+c\Bigg)\\
&\geq&8\Bigg(5\sqrt[5]{\frac1{2a^2b^2c}}\Bigg)\Bigg(5\sqrt[5]{\frac{a^2b^2c}{2^4}}\Bigg)\\
&=& 100
\end{eqnarray}
$$
Hence,
$$
\begin{eqnarray}
k &\leq& \frac{(a+b)^2+(a+b+4c)^2}{abc}(a+b+c)\\
&\leq& 100\tag*{$\square$}
\end{eqnarray}
$$
