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.

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 $y=\frac{1}{2}\sqrt{1+\frac{1}{\sqrt{5}}}$.
