Showing $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$

Question

Let $a$, $b$, $c$ be nonnegative real numbers.
Prove $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$.

Answer 1

I will prove the stronger inequality: $$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2$$ because $$(a^2+2)(b^2+2)=(a^2+1)(b^2+1)+a^2+b^2+3\ge (a+b)^2+\dfrac{1}{2}(a+b)^2+3 =\dfrac{3}{2}[(a+b)^2+2]$$ so $$(a^2+2)(b^2+2)(c^2+2)\ge \dfrac{3}{2}[(a+b)^2+2](c^2+2)\ge\dfrac{3}{2} [\sqrt{2}(a+b)+\sqrt{2}c]^2=3(a+b+c)^2$$ so $$(a^2+2)(b^2+2)(c^2+2)\ge 3(a+b+c)^2\ge 9(ab+bc+ac)$$

Answer 2

Expanding, we have $$ \left(a^2+2\right)\left(b^2+2\right)=2(a+b)^2+(ab-2)^2\tag1 $$ Substituting $b\mapsto b+c$, we get $$ \left(a^2+2\right)\left((b+c)^2+2\right)=2(a+b+c)^2+(a(b+c)-2)^2\tag2 $$ Substituting $(a,b)\mapsto(b,c)$ in $(1)$ gives $$ \begin{align} \left(b^2+2\right)\left(c^2+2\right) &=2(b+c)^2+(bc-2)^2\\ &=(b+c)^2+(b-c)^2+4bc+(bc-2)^2\tag{3a}\\ &=(b+c)^2+(b-c)^2+b^2c^2+4\tag{3b}\\ &=(b+c)^2+b^2+c^2-2bc+b^2c^2+4\tag{3c}\\ &=(b+c)^2+2+b^2+c^2+(bc-1)^2+1\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: $(b+c)^2=(b-c)^2+4bc$
$\text{(3b)}$: $4bc+(bc-2)^2=b^2c^2+4$
$\text{(3c)}$: $(b-c)^2=b^2+c^2-2bc$
$\text{(3d)}$: $b^2c^2-2bc+1=(bc-1)^2$

Multiply $(3)$ by $a^2+2$: $$ \begin{align} &\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\\ &=\left(a^2+2\right)\left((b+c)^2+2\right)+\left(a^2+2\right)\left(b^2+c^2+(bc-1)^2+1\right)\\ &=2(a+b+c)^2\color{#AAA}{+(a(b+c)-2)^2}+\left(a^2+2\right)\left(b^2+c^2\color{#AAA}{+(bc-1)^2}+1\right)\tag{4a}\\ &=2(a+b+c)^2+\left(a^2+2\right)\left(b^2+c^2+1\right)\\ &\:\color{#AAA}{+\left(a^2+2\right)(bc-1)^2+(a(b+c)-2)^2}\tag{4b}\\ &=2(a+b+c)^2+a^2+b^2+c^2+b^2+c^2+a^2b^2+1+a^2c^2+1\\ &\:\color{#AAA}{+\left(a^2+2\right)(bc-1)^2+(a(b+c)-2)^2}\tag{4c}\\ &=2(a+b+c)^2+a^2+b^2+c^2\\ &+2bc\color{#AAA}{+(b-c)^2}+2ab\color{#AAA}{+(ab-1)^2}+2ac\color{#AAA}{+(ac-1)^2}\\ &\:\color{#AAA}{+\left(a^2+2\right)(bc-1)^2+(a(b+c)-2)^2}\tag{4d}\\ &=3(a+b+c)^2\\ &\:\color{#AAA}{+\,(b-c)^2+(ab-1)^2+(ac-1)^2+\left(a^2+2\right)(bc-1)^2+(a(b+c)-2)^2}\tag{4e} \end{align} $$ Explanation:
$\text{(4a)}$: apply $(2)$
$\text{(4b)}$: rearrange terms
$\text{(4c)}$: expand product
$\text{(4d)}$: complete squares
$\text{(4e)}$: $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Thus, $$ \left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\ge3(a+b+c)^2\tag5 $$ and equality only happens when $a=b=c=\pm1$.

Therefore, $$ \begin{align} &\left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\\ &\ge3(a+b+c)^2\tag{6a}\\ &=6(ab+bc+ca)+3\!\left(a^2+b^2+c^2\right)\tag{6b}\\ &=6(ab+bc+ca)\\ &\:+3\!\left(ab\color{#AAA}{+\tfrac12(a-b)^2}+bc\color{#AAA}{+\tfrac12(b-c)^2}+ca\color{#AAA}{+\tfrac12(c-a)^2}\right)\tag{6c}\\ &=9(ab+bc+ca)\color{#AAA}{+\tfrac32\left((a-b)^2+(b-c)^2+(c-a)^2\right)}\tag{6d} \end{align} $$ Explanation:
$\text{(6a)}$: $(5)$
$\text{(6b)}$: expand square
$\text{(6c)}$: complete squares
$\text{(6d)}$: collect terms

Thus, $$ \left(a^2+2\right)\left(b^2+2\right)\left(c^2+2\right)\ge9(ab+bc+ca)\tag7 $$ where equality happens only when $a=b=c=\pm1$.

Note that if equality happens in $(5)$, then equality happens in $(7)$ (the converse is obvious).