<s>[INST] Prove $\sum ab \sum \frac{1}{(a+b)^2} \geqslant \frac{9}{4}+\frac{kabc\sum (a^2-bc)}{(a+b+c)^3(ab+bc+ca)}$ for the best k. For $a,b,c\geqslant 0;ab+bc+ca>0.$ Find $k_\max$ and proving in that case$:$
$$(ab+bc+ca)\Big(\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(c+a)^2}\Big) \geqslant \dfrac{9}{4}+\dfrac{kabc(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^3(ab+bc+ca)}.$$
I use computer and I found $k_\max =4.$ Then proving$:$
$$(ab+bc+ca)\Big(\dfrac{1}{(a+b)^2}+\dfrac{1}{(b+c)^2}+\dfrac{1}{(c+a)^2}\Big) \geqslant \dfrac{9}{4}+\dfrac{4abc(a^2+b^2+c^2-ab-bc-ca)}{(a+b+c)^3(ab+bc+ca)}.$$
Let $p=a+b+c,q=ab+bc+ca, r=abc,$ need to prove$:$
$$f(r)=(48q  -16p^2) r^3+pq ( 23{p}^{2}-96q
) {r}^{2}+6{p}^{2}{q}^{2} ( 3{p}^{2}+8q ) r+{p
}^{3}{q}^{2} \left( {p}^{2}-4\,q \right)  \left( 4{p}^{2}-q
 \right)  \geqslant 0.$$
Since $$f'(r)=3 \left( 48q -16\,{p}^{2}\right) {r}^{2}+2pq ( 23\,{p}^{2}
-96\,q ) r+6\,{p}^{2}{q}^{2} \left( 3\,{p}^{2}+8\,q \right) 
\geqslant 0$$
$$\because (a-b)^2 (b-c)^2 (c-a)^2 \geqslant 0.$$
Therefore we need to prove $f(r)\geqslant 0$ when $r$ get the minimum values.
Let $p=1,q=\dfrac{1-t^2}{3} (0 \leqslant t \leqslant 1),r=abc$ we have$:$
$$\dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2} \leqslant r$$
(see the proof here)
So $$f(r) \geqslant f\Big(\dfrac{1}{27} \left( 1-2t \right) \left( 1+t \right) ^{2}\Big)=$$
$$={\frac {8}{19683}}\,{t}^{2} \left( 4\,{t}^{3}-6\,{t}^{2}-15\,t+49
 \right)  \left( t-2 \right) ^{2} \left( 2\,t-1 \right) ^{2} \left( t+
1 \right) ^{2}
 \geqslant 0,$$
which is true.
But I can't prove $f'(r) \geqslant 0$ with $(a-b)^2(b-c)^2(c-a)^2\geqslant 0.$
I only check it with computer and know it's true because $\prod (a-b)^2 \geqslant 0.$
In addition$,$ I wish to know how do you find $k_\max=4$ if without computer$?$
Source. I post it to found a proof for this inequality, there was a SOS's proof.
 [/INST] Find $k_{\max}.$ Let $a=b=1$ then
$$(2c+1)\left[\frac 14+\frac{2}{(c+1)^2}\right] \geqslant \frac 94+\frac{k \cdot c(c-1)^2}{(c+2)^3(2c+1)},$$
equivalent to
$$\frac{c(c-1)^2}{2(c+1)^2}\geqslant \frac{k \cdot c(c-1)^2}{(c+2)^3(2c+1)},$$
or
$$c(c-1)^2\left[k - \frac{(2c+1)(c+2)^3}{2(c+1)^2}\right] \leqslant 0.$$
Therefore, we will to find $k$ satisfy
$$k \leqslant \frac{(2c+1)(c+2)^3}{2(c+1)^2}.$$
Let $c=0,$ we get $k \leqslant 4.$
</s>