An inequality involving real numbersLet x, y, z be real numbers such that&nbsp;xyz=−1. Prove that3232≥E:=4(x3+y3+z3)(x2+y2+z2)2

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An inequality involving real numbersLet x, y, z be real numbers such that&amp;nbsp;xyz=−1. Prove that3232≥E:=4(x3+y3+z3)(x2+y2+z2)2

Jeremiah Dickerson · Accepted Answer

Step 1Let&amp;nbsp;x+y+z=3u,&amp;nbsp;xy+xz+yz=3v2,, where&amp;nbsp;v2&amp;nbsp;can be negative, and&amp;nbsp;xyz=w3Thus, we need to prove that:(32)23(x2+y2+z2)2≥-4xyz3(x3+y3+z3).Now, we can assumme that&amp;nbsp;x3+y3+z3&amp;gt;0,, otherwise the inequality is obviously true.But, if sox3+y3+z3−3xyz&amp;gt;0or(x+y+z)∑cyc(x−y)2&amp;gt;0,which gives&amp;nbsp;u&amp;gt;0Id est, we need to prove that:3(32)23(3u2−2v2)2≥−4w(9u3−9uv2+w3)or12(32)23v4−(36uw+36(32)23u2)v2+27(32)23u4+36u3w+4w4≥0,for which it&#039;s enough to prove that324u2(w+(32)23u)2−12(32)23(27(32)23u4+36u3w+4w4)≤0or18(32)23u3w+27u2w2−4(32)23w4≤0or18(32)23u3+27u2w−4(32)23w3≥0,which is true by AM-GM:18(32)23u3−4(32)23w3+27u2w≥≥3(9(32)23u3)2·(-4(32)23w3)3+27u2w=0and we are done!

An inequality involving real numbers Let x, y, z

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