<s>[INST] $2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$ for real numbers $a,b,c$ $a,b,c$ 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:
  $$2(1+abc)+\sqrt2(1+a^2)(1+b^2)(1+c^2)\ge(1+a)(1+b)(1+c)$$

1With some help from Google translate
 [/INST] 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 
$$(u-v-w+1)^2\ge 0$$
</s>