Starting an inequality family: How far can we go?

Question

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.

Answer 1

Maybe the following my delirium will help.

It's enough to prove our inequality for all $k\geq1$.

A full expending gives $f(k)\geq0$, where $$f(k)=\sum_{cyc}(a^3b^3+a^3b^2c+a^3c^2b-3a^2b^2c^2)k^6+$$ $$+2\sum_{cyc}(a^4c^2+a^3b^3+a^4bc-7a^3b^2c+3a^3c^2b+a^2b^2c^2)k^5+$$ $$+\sum_{cyc}(a^5b+a^5c-9a^4b^2+4a^4c^2+2a^3b^3+5a^4bc+10a^3b^2c-22a^3c^2b+8a^2b^2c^2)k^4+$$ $$+2\sum_{cyc}(a^5b+a^5c+a^4b^2+a^4c^2-8a^3b^3-6a^4bc+8a^3b^2c+8a^3c^2b-6a^2b^2c^2)k^3+$$ $$+\sum_{cyc}(a^5b+a^5c+4a^4b^2-9a^4c^2+2a^3b^3+5a^4bc-22a^3b^2c+10a^3c^2b+8a^2b^2c^2)k^2+$$ $$+2\sum_{cyc}(a^4b^2+a^3b^3+a^4bc+3a^3b^2c-7a^3c^2b+a^2b^2c^2)k+$$ $$+\sum_{cyc}(a^3b^3+a^3b^2c+a^3c^2b-3a^2b^2c^2).$$ We'll prove that $$\sum_{cyc}(a^4c^2+a^3b^3+a^4bc-7a^3b^2c+3a^3c^2b+a^2b^2c^2)\geq0.$$ Indeed, we need to prove that $$\sum_{cyc}\left(\frac{a^2}{b^2}+\frac{ab}{c^2}+\frac{a^2}{bc}-\frac{7a}{c}+\frac{3a}{b}+1\right)\geq0.$$ Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.

Hence, $xyz=1$ and we need to prove that $$\sum_{cyc}(x^2+x^2y+x^2z-7xy+3x+1)\geq0.$$ Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, we need to prove that $$(9u^2-6v^2)w+9uv^2-3w^3-21v^2w+9uw^2+3w^3\geq0$$ or $g(v^2)\geq0$, where $$g(v^2)=(u-3w)v^2+u^2w+uw^2.$$ We see that $g$ is a linear function, which says that it's enough to prove the last inequality for an extremal value of $v^2$, which happens for equality case of two variables.

Let $y=x$ and $z=\frac{1}{x^2}$.

We need to prove that $$(x-1)^2(2x^5-x^4+2x^3+10x^2+4x+1)\geq0,$$ which is obvious.

Hence, $$f''''(k)=360\sum_{cyc}(a^3b^3+a^3b^2c+a^3c^2b-3a^2b^2c^2)k^2+$$ $$+240\sum_{cyc}(a^4c^2+a^3b^3+a^4bc-7a^3b^2c+3a^3c^2b+a^2b^2c^2)k+$$ $$+24\sum_{cyc}(a^5b+a^5c-9a^4b^2+4a^4c^2+2a^3b^3+5a^4bc+10a^3b^2c-22a^3c^2b+8a^2b^2c^2)\geq$$ $$\geq360\sum_{cyc}(a^3b^3+a^3b^2c+a^3c^2b-3a^2b^2c^2)+$$ $$+240\sum_{cyc}(a^4c^2+a^3b^3+a^4bc-7a^3b^2c+3a^3c^2b+a^2b^2c^2)+$$ $$+24\sum_{cyc}(a^5b+a^5c-9a^4b^2+4a^4c^2+2a^3b^3+5a^4bc+10a^3b^2c-22a^3c^2b+8a^2b^2c^2)=$$ $$=24\sum_{cyc}(a^5b+a^5c+4a^4b^2+a^4c^2+27a^3b^3+15a^4bc-45a^3b^2c+23a^3c^2b-27a^2b^2c^2)\geq0.$$ Hence, $$f'''(k)=120\sum_{cyc}(a^3b^3+a^3b^2c+a^3c^2b-3a^2b^2c^2)k^3+$$ $$+120\sum_{cyc}(a^4c^2+a^3b^3+a^4bc-7a^3b^2c+3a^3c^2b+a^2b^2c^2)k^2+$$ $$+24\sum_{cyc}(a^5b+a^5c-9a^4b^2+4a^4c^2+2a^3b^3+5a^4bc+10a^3b^2c-22a^3c^2b+8a^2b^2c^2)k+$$ $$+12\sum_{cyc}(a^5b+a^5c+a^4b^2+a^4c^2-8a^3b^3-6a^4bc+8a^3b^2c+8a^3c^2b-6a^2b^2c^2)\geq$$ $$\geq120\sum_{cyc}(a^3b^3+a^3b^2c+a^3c^2b-3a^2b^2c^2)+$$ $$+120\sum_{cyc}(a^4c^2+a^3b^3+a^4bc-7a^3b^2c+3a^3c^2b+a^2b^2c^2)+$$ $$+24\sum_{cyc}(a^5b+a^5c-9a^4b^2+4a^4c^2+2a^3b^3+5a^4bc+10a^3b^2c-22a^3c^2b+8a^2b^2c^2)+$$ $$+12\sum_{cyc}(a^5b+a^5c+a^4b^2+a^4c^2-8a^3b^3-6a^4bc+8a^3b^2c+8a^3c^2b-6a^2b^2c^2)=$$ $$12\sum_{cyc}(3a^5b+3a^5c-17a^4b^2+19a^4c^2+16a^3b^3+14a^4bc-32a^3b^2c+4a^3c^2b-10a^2b^2c^2)\geq$$ $$12\sum_{cyc}(3a^5b+3a^5c-17a^4b^2+19a^4c^2+14a^3b^3+14a^4bc-30a^3b^2c+4a^3c^2b-10a^2b^2c^2)=$$ $$=12\sum_{cyc}(3x^3y+3x^3z-17x^2y^2+19x^2+14x^2y+14x^2z-30xy+4x-10),$$ which can be negative!

Answer 2

"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.

Answer 3

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 $0\,$.