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

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?

• $$(a+b)^2(c+d)^2-16abcd=\left((a+b)(c+d)-8\frac{bcd}{c+d}\right)^2+16b^2cd\left(\frac{c-d}{c+d}\right)^2$$ – Did Feb 3 '15 at 14:56
• @Did. This is exactly what I asked. Than you very much. – user164524 Feb 3 '15 at 17:45
• @Did: Please consider making your comment into an answer, so the OP can accept it, and this post can move off the "Unanswered" queue. Thanks! – Kieren MacMillan Sep 3 '17 at 20:09
• @KierenMacMillan Did you note the OP's account is inactivated? – Did Sep 3 '17 at 20:50
• @Did: No… Good catch. So how do we get this off the queue? Close? – Kieren MacMillan Sep 3 '17 at 22:00