Reduce $(a+b)^2(c+d)^2-16abcd$ to sum of squares

Question

Is it possible to reduce $$(a+b)^2(c+d)^2-16abcd$$ to sum of squares?
This expression is used in proof of AM-GM inequality. It is known that $$\dfrac{a+b}{2}\ge\sqrt{ab}\tag{1}$$ So, it can be proved reducing to sum of squares (in this case only one square): $$a+b\ge2\sqrt{ab}\\ a-2\sqrt{ab}+b\ge0\\ \left(\sqrt a-\sqrt b\right)^2\ge0 $$ which is obviously true. So, we can say that for all non-negative real numbers $a,b$ it is true that $$\dfrac{a+b}2\ge\sqrt{ab}$$ If we want to prove that $$\dfrac{a+b+c+d}{4}\ge\sqrt[4]{abcd}$$ we will write $$\sqrt[4]{abcd}=\sqrt{\sqrt{ab}\sqrt{cd}}\le\sqrt{\dfrac{a+b}2\cdot\dfrac{c+d}{2}}\le\dfrac{a+b+c+d}{4}$$ There is only used $(1)$ to prove this inequality (in this case we used it $2$ times). It is known that $(1)$ can be reduced to sum of squares. Then, is it possible to reduce $$\sqrt{\sqrt{ab}\sqrt{cd}}\le\sqrt{\dfrac{a+b}2\cdot\dfrac{c+d}{2}}$$ to sum of squares? It is same as $(a+b)^2(c+d)^2-16abcd$, but what next? If it is impossible, can we prove that?