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Inequality

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Grigorios Kostakos
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Inequality

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Post by Grigorios Kostakos » Fri May 26, 2017 3:52 am

Let $a,b,c$, and $d$ be strictly positive real numbers such that the arithmetic mean of the six numbers $ab,\, ac,\, ad,\, bc,\, bd,\, cd$ equals to the square of their geometric mean. Prove that $$a+b+c+d - 2\sqrt{2}\,abcd\leqslant 4 - 2\sqrt{2}\,.$$
When is the equality attained?
Grigorios Kostakos
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