Starting an inequality family: How far can we go? Consider the inequality
$$(ab+bc+ca)\cdot\left(\frac{1}{(ka+b)^2}+\frac{1}{(kb+c)^2}+\frac{1}{(kc+a)^2}\right)\,\ge\,\frac{9}{(k+1)^2}\tag{"Case $k$"}$$
with variables $\,a,b,c\in\mathbb{R}^{>0}\,$ and parameter $\,k\in\mathbb{R}$.
(At least) the two instances with $\,k=1\,$ and
$\,k=2\,$
have yet found their home at math.SE being answered in the positive. The case $\,k=1\,$ is entitled 
'Hard inequality' (aka "Iran 1996" amongst insiders I guess), cf the comments there containing further references.
My question: For which other values of the parameter $k$ does the inequality "Case $k$" hold true?
Please note that "Case $k$" is invariant under replacing $\,k\mapsto \frac{1}{k}\,$ and simultaneously switching any two out of the three variables.
So I'd expect that any $\,k>0\,$ yields a valid statement. To 'complete the proof job' it would suffice if a reduction from $\,k>1\,$to $\,k=1\,$ can be achieved.
 A: "Case k=0"
Just another brick in the wall, not a comprehensive answer. It is shown that the inequality holds true for the parameter value $k=0$.
After clearing denominators we face the expression
$$(ab+bc+ca)\left[\sum_\text{cyc}a^2b^2\right]\;-\;9a^2b^2c^2$$
to be transformed in a sum of squares with positive prefactors.
In the first summand separate terms involving all three variables from those containing two, and spread the multiples of $a^2b^2c^2$ accordingly
$$\begin{eqnarray}
 & =\quad & abc\left[\sum_\text{cyc}\left(b^2c+bc^2\right)-6abc\right]\;+\;\underbrace{\sum_\text{cyc}a^3b^3 - 3a^2b^2c^2}_\text{is AM-GM} \\[3ex]
 & =\quad & abc\sum_\text{cyc}a(b-c)^2\:
 +\:\frac{ab+bc+ca}{2}\,\sum_\text{cyc}a^2(b-c)^2
\end{eqnarray}$$
To the underbraced summand the identity
$r^3+s^3+t^3-3rst=\frac 12(r+s+t)\sum_\text{cyc}(r-s)^2$
with $ab,bc,ca\,$ inserted has been applied.
A: A proof in the parameter range $\,k\geqslant 0\,$ is proposed which relies on the known case $k=1$.
It encompasses the following steps, hereby submitted to the community's critical eye:


*

*Variable transformation, to simplify the subsequent

*clearing of denominators.

*Scaling the inequality and

*arguing that its $k=1$ instance yields a lower bound for other values of $k$.

*Limit case $k\to 0$


Let's send $\:a,b,c\:$ to the new variables $\:u,v,w\:$ via
$$\begin{pmatrix}u\\ v\\ w\end{pmatrix}\;=\; 
\begin{pmatrix}k&1&0\\0&k&1\\1&0&k\end{pmatrix}\,
\begin{pmatrix}a\\b\\c\end{pmatrix}$$
Note that $\,u+v+w=(k+1)(a+b+c)$, and especially
$$\begin{matrix}
\sum_\text{cyc}uv & = & \left(k^2+k+1\right)\sum_\text{cyc}ab & +
 & k\:\sum_\text{cyc}a^2 \\[1.5ex]
\sum_\text{cyc}u^2 & = & 2k\:\sum_\text{cyc}ab &
 + & \left(k^2+1\right)\sum_\text{cyc}a^2\,.
\end{matrix}$$
Assume $\,k>0\,$ and combine the preceding two identities to obtain
$$
\left(k+\frac 1 k +1\right)(k+1)^2\sum_\text{cyc}ab\;=\;
\left(k+\frac 1 k\right)\sum_\text{cyc}uv -\sum_\text{cyc}u^2\,.
$$
This is used when multiplying ("Case $k$"), the given inequality, with $\,(k+\frac 1 k -1)(k+1)^2\,u^2v^2w^2$, and after clearing the RHS let
$$
g(k,u,v,w)\;:=\;\left[\left(k+\frac 1 k\right)\sum_\text{cyc}uv -\sum_\text{cyc}u^2 \right]
\cdot\sum_\text{cyc}u^2v^2\; -\;9\left(k+\frac 1 k -1\right)u^2v^2w^2\,.
$$
Note that the freedom to scale the function $g$ has been exploited such that the second "$\sum$" in the brackets does not depend on $k\;\implies$ will drop out upon differentiation w.r.to $k$.
It has to be shown that $\,g(k,u,v,w)\geq 0\;\:\forall\, k,u,v,w>0\,$.
This is accomplished by one-variable analysis in $k$ with $\,u,v,w\,$ fixed.
$g$ enjoys the invariance property $\,g(1/k,\ldots)=g(k,\ldots)\,$ implying
$\,\frac 1kg_k\big(\frac 1k,\ldots\big)=-k\,g_k(k,\ldots)\,$ where $g_k\equiv\frac{\partial}{\partial k}g$. In particular $\,g_k(1,\ldots)=0\,$. Thus the known case $\,g(1,\ldots)\geqslant0\,$ is a bound for $\,g(k,\ldots)\,$, but we badly need $\,g_k(k,\ldots)\geqslant0\,$ for $\,k\geqslant 1\,$ to conclude it's a lower bound!
Now
$$
g_k(k,u,v,w)\;=\;\left(1-\frac 1{k^2}\right)
\left[\left(\sum_\text{cyc}uv\right)\left(\sum_\text{cyc}u^2v^2\right)\; -\;9u^2v^2w^2\right]
$$
and separation of terms involving all three variables from those containing two and distributing the multiples of $\,u^2v^2w^2\,$ accordingly leads to
$$
=\;\left(1-\frac 1{k^2}\right)
\left[uvw\sum_\text{cyc}u(v-w)^2\; +\;\frac{uv+vw+wu}{2} 
\sum_\text{cyc}u^2(v-w)^2\right]\,.
$$
Bingo, coz the bracket is a sum of squares with positive coefficients. This completes the $\,k>0\,$ case.
Since the inequality is built on "$\le$" (and not "$<$") the case "$k=0$" holds by continuity too, just send $k$ to $0\,$.
