How to prove this inequality? $ab+ac+ad+bc+bd+cd\le a+b+c+d+2abcd$ 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.
 A: Let $a=\sqrt 3\cos x\cos y,b=\sqrt 3\cos x\sin y,c=\sqrt 3\sin x\cos y,d=\sqrt 3\sin x\sin y, \pi/2>x,y>0$.
A: Your constraints are $a,b,c,d > 0$ and $a^2+b^2+c^2+d^2 = 3$, by doing a change of variables, $x^2 = a^2/3, y^2 = b^2/3, z^2 = c^2/3, w^2 = d^2/3$. Then your constraints change to 
$$x,y,z,w > 0, x^2+y^2+z^2+w^2 =1$$
and the problem is equivalent to proving $xy +xz +xw +yz+yw+zw \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$. Since $x^2+y^2+z^2+w^2 =1$ that first implies that $x,y,z,w \leq 1$ and in particular $x^2+y^2+z^2+w^2 \leq x+y+z+w$. Doing your standard expansion,
$$ (x+y+z+w)^2 = x^2+y^2+z^2+w^2 + 2(xy+xz+xw+yz+yw+zw) $$
The right hand side is bounded above by $x^2+y^2+z^2+w^2 + (x^2+z^2)+(x^2+w^2)+(y^2+z^2) +(y^2 + w^2) + 2(xy + wz) = 3(x^2+y^2+z^2+w^2) + 2(xy+wz)$
Balancing this with the RHS of the display we get 
$$2(xy+xz+xw+yz+yw+zw) \leq 2(x^2+y^2+z^2+w^2) + 2(xy+wz)$$
Then we have that $xy \leq (x^2+y^2)/2$ and $wz \leq (w^2+z^2)/2$ and placing this in we have 
$$xy+xz+xw+yz+yw+zw \leq 1.5(x^2+y^2+z^2+w^2) \leq \sqrt{3}(x+y+z+w) + \frac{2}{3}xyzw$$
since $\frac{2}{3}xyzw \geq 0$, $1.5 <\sqrt{3}$ and $x^2+y^2+z^2+w^2 \leq x+y+z+w$.
A: In addition: Under the same constraints, does the inequality
$$ab+ac+ad+bc+bd+cd\le a+b+c+d+kabcd$$
always hold, for $k<2?$
