Proving $\displaystyle \sum \dfrac{(a+b)(a+c)}{(b-c)^2} \geqslant 3+4\sqrt{2}$

Problem. For $a,b,c>0.$ Prove: $$  \dfrac{(a+b)(a+c)}{(b-c)^2}+\dfrac{(b+a)(b+c)}{(a-c)^2} +\dfrac{(c+a)(c+b)}{(a-b)^2} \geqslant 3+4\sqrt{2}$$

Solution. (tthnew) $\text{The inequality is equivalent to}$

$$\sum f(a,b,c) (-{a}^{2}+ab+\sqrt {2}ab+ac-\sqrt {2}ca-{b}^{2}+bc-\sqrt {2}cb-{c}^{2}+ \sqrt {2}{c}^{2})^2 \geqslant 0,\, \text{where}$$

$$f(a,b,c) = \left( \dfrac{\sqrt{2}}{17}+{\dfrac {16}{51}} \right) {a}^{2}+ \left( \dfrac{\sqrt{2}}{17} +{\dfrac {16}{51}} \right) {b}^{2}+ \left( {\frac {11}{17}}+ {\frac {20}{51}}\,\sqrt {2} \right) {c}^{2}+ $$

$$+\left( {\frac {55}{51}}+{ \frac {56}{51}}\,\sqrt {2} \right) ab+ \left( -{\frac {10}{17}}-{ \frac {29}{51}}\,\sqrt {2} \right) ac+ \left( -{\frac {10}{17}}-{ \frac {29}{51}}\,\sqrt {2} \right) bc$$

$\text{But}$ $$\sum f(a,b,c) ={\dfrac {2\sqrt {2}+5 }{51}}  \left( 13\,{a}^{2}+13 \,{b}^{2}+13\,{c}^{2}-ab-ac-bc \right) \geqslant 0, \, \text{and}$$

$$\sum f(a,b,c) \cdot f(b,c,a) \geqslant 0 \, (\text{easy prove!})$$

$\text{So we are done.} \square$