Let a, b, c, d be positive numbers. Prove that (a2+b2+c2+d2)2≥(a+b)(b+c)(c+d)(d+a). When does equality hold?
Solution
By the inequality between the arithmetic and the geometric mean, we have (a+b)(b+c)(c+d)(d+a)≤(4(a+b)+(b+c)+(c+d)+(d+a))4=24(4a+b+c+d)4. By the inequality between the quadratic and the arithmetic mean, we have 24(4a+b+c+d)4≤24(4a2+b2+c2+d2)2=(a2+b2+c2+d2)2, as required. In the second inequality, equality holds if and only if a=b=c=d, but in that case, equality holds also in the original inequality. Therefore, equality holds if and only if a=b=c=d.
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