Mathematical Reflections 2017, Issue 1 - Problem J398

Problem:
Let $a,b,c$ be real numbers. Prove that $$(a^2+b^2+c^2-2)(a+b+c)^2+(1+ab+bc+ca)^2 \geq 0.$$

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
Let $x=a+b+c$ and $y=ab+bc+ca$. Then, $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=x^2-2y.$$ So,
$$\begin{array}{lll} (a^2+b^2+c^2-2)(a+b+c)^2+(1+ab+bc+ca)^2&=&(x^2-2y-2)x^2+(1+y)^2\\&=&x^4-2x^2(y+1)+(y+1)^2\\&=&(x^2-y-1)^2 \\ & \geq & 0. \end{array}$$
Equality holds if and only if $x^2=y+1$, i.e. if and only if $(a+b+c)^2=ab+bc+ca+1$, i.e. if and only if $a^2+b^2+c^2=1-ab-bc-ca$.