Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$ For $a,b,c>0.$ Prove$:$ $$4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}$$
My proof is using SOS$:$
$${c}^{2}{a}^{2} {b}^{2}\Big( \sum a\Big)^2 \sum a^2  \Big\{ 4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}-{\dfrac {7(a+b+c)}{abc}} \Big\}$$
$$=\dfrac{1}{2} \sum {a}^{2}{b}^{2} \left( {a}^{2}+{b}^{2}-2\,{c}^{2} +5bc-10ab+5\, ac \right) ^{2} +\dfrac{1}{2} \prod (a-b)^2 \left( 7\sum a^2 +50\sum bc \right)  \geqslant 0.$$
From this we see that the inequality is true for all $a,b,c \in \mathbb{R};ab+bc+ca\geqslant 0.$
But we also have this inequality for $a,b,c \in \mathbb{R}.$ Which verify by Maple.
I try and I found a proof but I'm not sure$:$
If replace $(a,b,c)$ by $(-a,-b,-c)$ we get the same inequality.
So we may assume $a+b+c\geqslant 0$ (because if $a+b+c<0$ we can let $a=-x,b=-y,c=-z$ where $x+y+z \geqslant 0$ and the inequality is same!)
Let $a+b+c=1,ab+bc+ca=\dfrac{1-t^2}{3} \quad (t\geqslant 0), r=abc.$ Need to prove$:$
$$f(r) =81\,{r}^{2}-15\,r+\dfrac{4}{9} \left( t-1 \right) ^{2} \left( t+1 \right) ^{2
}\geqslant 0.$$
It's easy to see, when $r$ increase then $f(r)$ decrease. Since $r\leqslant \dfrac{1}{27} \left( 2\,t+1 \right) \left( t-1\right) ^{2} \quad$(see here).  We get$:$
$$f(r)\geqslant f\Big(\dfrac{1}{27} \left( 2\,t+1 \right) \left( t-1\right) ^{2}\Big)=\dfrac{1}{9} {t}^{2} \left( 2\,t-1 \right) ^{2} \left( t-1 \right) ^{2} \geqslant 0.$$
Done.
Could you check it for me? Who have a proof for $a,b,c \in \mathbb{R}$?
 A: For $a,\,b,\,c$ are real numbers. We have
$$(a+b+c)^2 =(|a+b+c|)^2 \leqslant (|a|+|b|+|c|)^2,$$
$$\frac{a+b+c}{abc}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \leqslant \left |\frac{1}{ab}  \right |+\left |\frac{1}{bc}  \right |+\left |\frac{1}{ca}  \right |=\frac{|a|+|b|+|c|}{|a||b||c|}.$$
So, we need to prove
$$4\left(\dfrac{1}{|a|^2}+\dfrac{1}{|b|^2}+\dfrac{1}{|c|^2} \right)+\dfrac{81}{(|a|+|b+|c|)^2} \geqslant \frac{7(|a|+|b|+|c|)}{|a||b||c|}.$$
Now, replace $(|a|,|b|,|c|) \to (a,b,c)$ then $a,b,c \geqslant 0.$ The inequality become
$$4\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}.$$
This is the original inequality.
A: After using the nguyenhuyen_ag's reasoning it's enough to prove our inequality for positive variables,
and we can end the proof by $uvw$.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that:
$$\frac{4(9v^4-6uw^3)}{w^6}+\frac{81}{9u^2}\geq\frac{21u}{w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=w^6-5u^3w^3+4u^2v^4.$$
But $$f'(w^3)=2w^2-5u^3<0,$$ which says that $f$ decreases and it's enough to prove our inequality for a maximal value of $w^3$,
which by $uvw$ happens for equality case of two variables.
Since our inequality is homogeneous and symmetric, it's enough to assume $b=c=1,$ which gives
$$(a-1)^2(a-4)^2\geq0$$ and we are done!
