Finding the maximum value of $ab+ac+ad+bc+bd+3cd$ 
If $a,b,c,d>0$ satisfy the condition ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }=1$, find the maximum value of $ab+ac+ad+bc+bd+3cd$.

I'm not progress in this inequality problem. Please help.
Thank you.
 A: Let $ab+ac+ad+bc+bd+3cd=k$.
Hence, $k>0$ and $ab+ac+ad+bc+bd+3cd=k(a^2+b^2+c^2+d^2)$ or 
$ka^2-(b+c+d)a+k(b^2+c^2+d^2)-bc-bd-3cd=0$.
Hence, $(b+c+d)^2-4k(k(b^2+c^2+d^2)-bc-bd-3cd)\geq0$ or
$(4k^2-1)b^2-2(2k+1)(c+d)b+(4k^2-1)(c^2+d^2)-2(6k+1)cd\leq0$.
If $0<k\leq\frac{1}{2}$ so the last inequality is obviously true.
Let $k>\frac{1}{2}$. Hence, $(2k+1)^2(c+d)^2-(4k^2-1)\left((4k^2-1)(c^2+d^2)-2(6k+1)cd\right)\geq0$ or
$(2k+1)(c+d)^2-(2k-1)\left((4k^2-1)(c^2+d^2)-2(6k+1)cd\right)\geq0$ or
$(2k^2-k-1)c^2-(6k-1)cd+(2k^2-k-1)d^2\leq0$.
If $\frac{1}{2}<k\leq1$ so the last inequality is obviously true.
Let $k>1$. Hence, $(6k-1)^2-4(2k^2-k-1)^2\geq0$, which gives $k\leq1+\frac{\sqrt5}{2}$.
Easy to see that for $k=1+\frac{\sqrt5}{2}$ the equality indeed occurs.
Id est, the answer is $1+\frac{\sqrt5}{2}$.
A: Hint: consider that the eigenvalues of the symmetric matrix
$$ M = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 3 \\ 1 & 1 & 3 & 0 \end{pmatrix} $$
are $-3,-1,2+\sqrt{5}$ and $2-\sqrt{5}$. In particular, $\left(\frac{\sqrt{5}-1}{2},\frac{\sqrt{5}-1}{2},1,1\right)^T$ in the only eigenvector of $M$ with positive coordinates, and it is associated with the largest eigenvalue $2+\sqrt{5}$. It follows that:

$$ \max_{\substack{a^2+b^2+c^2+d^2=1 \\ a,b,c,d>0}} \frac{1}{2}(a\, b\, c\, d)\, M\, (a\, b\, c\, d)^T = \frac{2+\sqrt{5}}{2} = \color{red}{1+\frac{\sqrt{5}}{2}}. $$

Such a maximum is attained by $(a,b,c,d)=(x,x,y,y)$ with $x=\frac{1}{2}\sqrt{1-\frac{1}{\sqrt{5}}}$ and $y=\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{5}}}$.
