Prove $P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$ For $a,b,c$ are reals$.$ Prove$:$ $$P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$$
I found this from Michael Rozenberg's solution. See here.
My proof:
$$P=\frac{1}{16} \, \left( a+b \right) ^{2} \left( a+b-4\,c \right) ^{2}+{\frac {5 \, \left( a+b \right) ^{4}}{14}}$$ $$+{\frac { \left( 3\,{a}^{2}+6\,ab+3\,{ b}^{2}-28\,{c}^{2} \right) ^{2}}{112}}+\frac{3}{8}\, \left( a+b \right) ^{2} \left( a-b \right) ^{2}+\frac{1}{8}\, \left( 2\,c+a+b \right) ^{2} \left( a-b \right) ^{2}$$
I’m looking for an alternative proof. Thanks!

Let $a+b=2u$ and $ab=v^2$, where $v^2$ can be negative.
Thus, we need to prove that:
$$7c^4-2abc^2-2ab(a+b)c+(a+b)^2(a^2+b^2)\geq0$$ or
$$7c^4-2v^2c^2-4uv^2c+8u^2(2u^2-v^2)\geq0$$ or
$$7c^4+16u^4\geq2v^2(c^2+2uc+4u^2).$$
But, $$c^2+2uc+4u^2=(c+u)^2+3u^2\geq0$$ and $v^2\leq u^2$ it's just $(a-b)^2\geq0.$
Thus, it's enough to prove that
$$7c^4+16u^4\geq2u^2(c^2+2uc+4u^2)$$ or
$$7c^4-2u^2c^2-4u^3c+8u^4\geq0,$$ which is true by AM-GM:
$$7c^4-2u^2c^2-4u^3c+8u^4\geq$$
$$\geq c^4+u^4-2c^2u^2+c^4+3u^4-4u^3c\geq$$
$$\geq4\sqrt[4]{c^4(u^4)^3}-4u^3c=4|u^3c|-4u^3c\geq0.$$
