show that $a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$

Question

let $a,b,c>0$. Show that $$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c)$$

By the $$(1+a)(1+b)(1+c)=1+a+b+c+ab+bc+ac+abc$$ it is suffient to show $$a^2+b^2+c^2+abc+2\ge a+b+c+ab+bc+ac$$ How?

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $$a^2+b^2+c^2+2+abc\geq a+b+c+ab+ac+bc$$ or $f(v^2)\geq0$, where $$f(v^2)=9u^2-6v^2+w^3+2-3u-3v^2.$$

But $f$ is a linear function,

which says that it's enough to prove our inequality foe an extremal value of $v^2$,

which happens for equality case of two variables.

Since our inequality is symmetric, we can assume $b=c$ and it's enough to prove that $$a^2+(b^2-2b-1)a+b^2-2b+2\geq0,$$ for which it's enough to prove that $$(b^2-2b-1)^2-4(b^2-2b+2)\leq0$$ for $b^2-2b-1\leq0$ or $$(b-1)^2(b^2-2b-7)\leq0,$$ which is obvious.

Done!

Another way.

Since $$\prod_{cyc}((a-1)(b-1))=\prod_{cyc}(a-1)^2\geq0,$$ we can assume that $$(a-1)(b-1)\geq0$$ or $$c(a-1)(b-1)\geq0$$ or $$abc\geq ac+bc-c.$$

Thus, it remains to prove that $$a^2+b^2+c^2+2+ac+bc-c\geq ab+ac+bc+a+b+c$$ or $$a^2+b^2+1-ab-a-b+(c-1)^2\geq0$$ or $$\frac{1}{2}((a-b)^2+(a-1)^2+(b-1)^2)+(c-1)^2\geq0.$$ Done again!

Answer 2

$$a,b,c \ge 0$$

$$a^2+b^2+c^2+2abc+3\ge (1+a)(1+b)(1+c) \Leftrightarrow $$

$$\Leftrightarrow (a-1)^2+(b-1)^2+(c-1)^2\ge (1-a)(1-b)(1-c)$$

Which is obviously.