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
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?
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,
$$\sqrt{2|(a-\sqrt{3}b)(c-\sqrt{3}d)|}+\sqrt{2|(a+\sqrt{3}b)(c+\sqrt{3}d)|}\geq\sqrt{2(|(a-\sqrt{3}b)(c-\sqrt{3}d)|+|(a+\sqrt{3}b)(c+\sqrt{3}d)|)}$$