$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$ for real numbers $a,b,c$

Question

$a,b,c$ reel sayılar için ;

$$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$$

Olduğunu gösteriniz.

Translation:¹ For real numbers $a,b,c$, show that: $$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$$

¹_{With some help from Google translate}

Answer 1

let $$u=a+b+c,v=ab+bc+ac,w=abc$$ it suffices to show that $$2(u^2+v^2+w^2-2wu-2v+1)\ge (u+v-w-1)^2$$ this inequality is equivalent to $$u^2+v^2+w^2-2uv+2vw-2uw+2u-2v-2w+1\ge 0$$ or $$(u-v-w+1)^2\ge 0$$