We know that x, y and Z are all positive numbers, and prove that 33 (LX + 1y + 1z) ≤ 1x2 + 1Y2 + 1z2

It is proved that (12 + 12 + 12) (1x2 + 1Y2 + 1z2) ≥ (1x + 1y + 1z) 2 is obtained from Cauchy inequality (5 points) then 3 × 1x2 + 1Y2 + 1z2 ≥ LX + 1y + 1z, that is 33 (LX + 1y + 1z) ≤ 1x2 + 1Y2 + 1z2 (10 points)

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