Proving inequality by SOS

$ 1.$ Prove:

$$ a^2 \geqslant \frac{(ab+bc+ca)(5a-b-c)}{3(a+b+c)}, \quad a > 0 , \; b > 0, \; c > 0.$$

Solution:

$$ \text{LHS}-\text{RHS} ={\frac { \left( ab+ca+bc \right) \left( 3\,a+b+c \right) \left( 2\,a-b-c \right) ^{2}+3\, \left( a+b+c \right) {a}^{2} \left( b-c \right) ^{2}}{3 \left( b+c \right) \left( 4\,a+b+c \right) \left( a+b+c \right) }}$$

$2.$ For $a,b,c \in \mathbb{R}.$ Prove$:$

$$ \displaystyle (a^2+b^2+c^2)^3 \geqslant 2(a-b)^2 (b-c)^2(c-a)^2$$

We have$ :$

$$ \displaystyle \text{LHS}-\text{RHS} ={\frac {2\, \prod \left( a+b-2\,c \right) ^{2}}{27}}+3\, \left( \sum a \right) ^{2} \left[ {\frac { (19 \sum a^2 -7 \sum ab)^2}{1539}}+{\frac { \left(\sum ab \right) ^{2}}{57 }} \right] \geqslant ~0$$