How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

Question

let $a,b,c,d\ge 0$,and $a^2+b^2+c^2+d^2=3$,prove that

$ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$

I find this inequality are same as Crux 3059 Problem.

Answer 1

Let $a=\sqrt 3\cos x\cos y,b=\sqrt 3\cos x\sin y,c=\sqrt 3\sin x\cos y,d=\sqrt 3\sin x\sin y, \pi/2>x,y>0$.

Answer 2

In addition: Under the same constraints, does the inequality $$ab+ac+ad+bc+bd+cd\le a+b+c+d+kabcd$$ always hold, for $k<2?$