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定理 设ａ，ｂ，ｃ为非负实数，记Ｐ＝∑ａ３＝ａ３＋ｂ３＋ｃ３，Ｑ＝∏ａ＝ａｂｃ，Ｒ＝∑ｂｃ（ｂ＋ｃ）＝ａ２ｂ＋ａｂ２＋ｂ２ｃ＋ｂｃ２＋ｃ２ａ＋ｃａ２，则 ２Ｐ≥Ｐ＋３Ｑ≥Ｒ≥６Ｑ．①证明：第一个不等式显然；由ａｂｃ≥（ｂ＋ｃ－ａ）（ｃ＋ａ－ｂ）（ａ＋ｂ－?
Theorem Let a, b, and c be non-negative real numbers, and remember that P=∑a3=a3+b3+c3,Q=∏a=abc,R=∑bc(b+c)=a2b+ab2+b2c+bc2+c2a+ca2, then 2P≥P+3Q≥R≥6Q. 1 Proof: The first inequality is obvious; from abc≥(b+c-a)(c+a-b)(a+b-?