Problem: Find the smallest number k such that for all real numbers x, y and z (x2+y2+z2)2≤k(x4+y4+z4)
Solution
Solution: Note that (x2+y2+z2)2=x4+y4+z4+2x2y2+2x2z2+2y2z2 Using the AM-GM Inequality, we find that 2x2y2+2x2z2+2y2z2≤2[(x4+y4)/2]+2[(x4+z4)/2]+2[(y4+z4)/2] which when combined with the previous equation results to (x2+y2+z2)2≤3(x4+y4+z4) Therefore k=3.
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Source: MathNet,
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