How prove this inequality for $a,b,c,d$ are real numbers let $a,b,c,d$ are real numbers,show that
$$2\sqrt{a^2+c^2}+\sqrt{a^2+c^2+3(b^2+d^2)-2\sqrt{3}(ab+cd)}+\sqrt{a^2+c^2+3(b^2+d^2)+2\sqrt{3}(ab+cd)}\ge6\sqrt{|ad-bc|}$$
This problem  is creat by  China's famous mathematician hua luogeng,http://en.wikipedia.org/wiki/Hua_Luogeng 
when he is child,and  Now this problem has some ugly methods,I think this inequality has nice methods,Thank you
 A: It is the isoperimetric inequality for a triangle with vertices $0$, $x+y$, and $2x$, where $x=(a,c)$ and $y = \sqrt{3} (b,d)$.  
If $P$ is the perimeter of the triangle and $A$ its area, Hua's inequality says $$|2x| + |x-y| + |x+y|  \geq 6\sqrt{\frac{A}{\sqrt{3}}} .$$ The sum on the left is $P$ and the squared inequality is $ P^2 \geq (12 \sqrt{3}) A$.  
This is the correct lower bound on $P^2/A$, because equality is attained at an equilateral triangle.    
In Hua's parameterization, the triangle is equilateral when $(a,c)$ is a 90 degree rotation of $(b,d)$, or $(a,c) = \pm (-d,b)$.  So it could be an inequality discovered by trying to prove the isoperimetric inequality (for triangles) in coordinates and choosing variables in which the algebra is simplified. Which is something that a very clever child might do.  But did he rediscover the isoperimetric principle on his own? 
A: Let $x=[a,c]^T$ and $y=\sqrt{3}[b,d]^T$ be two vector in the plane and denote the cross product by $\times$. The inequality is equivalent to $$2\Vert x\Vert+\Vert x+y\Vert+\Vert x-y\Vert\geq 2\sqrt{3\Vert x\times y\Vert}\iff\\4\Vert x\Vert^2+(\Vert x+y\Vert+\Vert x-y\Vert)^2+4\Vert x\Vert(\Vert x+y\Vert+\Vert x-y\Vert)\geq12\Vert x\times y\Vert.$$ For compactness let $u=x+y$ and $v=x-y$. Then we must show $$2\Vert u\Vert^2+2\Vert v\Vert^2+2\langle u,v\rangle+2\Vert u\Vert\Vert v\Vert+2\Vert u+v\Vert(\Vert u\Vert+\Vert v\Vert)\geq 6\Vert u\times v\Vert\iff\\\Vert u\Vert^2+\Vert v\Vert^2+\langle u,v\rangle+\Vert u\Vert\Vert v\Vert+\Vert u+v\Vert(\Vert u\Vert+\Vert v\Vert)\geq 3\Vert u\times v\Vert.$$ Note that $\Vert u+v\Vert\Vert u\Vert\geq\Vert (u+v)\times u\Vert=\Vert u\times v\Vert$, $\Vert u+v\Vert\Vert v\Vert\geq\Vert (u+v)\times v\Vert=\Vert u\times v\Vert$, and $\Vert u\Vert \Vert v\Vert+\langle u,v\rangle\geq0$. Therefore, it suffices to have $\Vert u\Vert^2 + \Vert v\Vert^2\geq\Vert u\times v\Vert$ which holds since the LHS is not less than $2\Vert u\Vert \Vert v\Vert$ which in turn is greater than or equal to $\Vert u\times v\Vert$.
An alternative geometric proof. Consider a triangle with vertices at $x$,$y$, and $-x$. The LHS is $2p$, the perimeter of the triangle, whereas the RHS is $2\sqrt{3S}$ with $S$ being the area of the triangle. Thus we have to show $p\geq\sqrt{3S}$. Raise both sides to four and use Heron's formula for $S^2$. We have to show $p^3\geq27(p-\alpha)(p-\beta)(p-\gamma)$ where $\alpha,\beta$, and $\gamma$ are the side lengths of the considered triangle. This last inequality is simply an AM-GM inequality.
A: HINT: 
$$2\sqrt{a^2+c^2}+\sqrt{a^2+c^2+3(b^2+d^2)-2\sqrt{3}(ab+cd)}+\sqrt{a^2+c^2+3(b^2+d^2)+2\sqrt{3}(ab+cd)}$$ $$=2\sqrt{a^2+c^2}+\sqrt{(a-\sqrt{3}b)^2+(c-\sqrt{3}d)^2}+\sqrt{(a+\sqrt{3}b)^2+(c+\sqrt{3}d)^2}$$ $$\geq 2\sqrt{2|ac|}+\sqrt{2|(a-\sqrt{3}b)(c-\sqrt{3}d)|}+\sqrt{2|(a+\sqrt{3}b)(c+\sqrt{3}d)|}$$
Now,
$$\sqrt{2|(a-\sqrt{3}b)(c-\sqrt{3}d)|}+\sqrt{2|(a+\sqrt{3}b)(c+\sqrt{3}d)|}\geq\sqrt{2(|(a-\sqrt{3}b)(c-\sqrt{3}d)|+|(a+\sqrt{3}b)(c+\sqrt{3}d)|)}$$ 
