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Algebra Difficulty 6.4 National olympiad Prove it

58. Let a,b,c,da, b, c, d be non-negative real numbers, A=a3+b3+c3+d3,B=bcd+cda+dab+abcA=a^{3}+b^{3}+c^{3}+d^{3}, B=b c d+c d a+d a b+a b c, prove: (a+b+c+d)3⩽4A+24B(a+b+c+d)^{3} \leqslant 4 A+24 B. (2000 Polish Mathematical Olympiad problem)

Solution

58. Let S=a+b+c+dS=a+b+c+d, then S3=A+6B+3QS^{3}=A+6 B+3 Q. Where
Q=a2(b+c+d)+b2(c+d+a)+c2(d+a+b)+d2(a+b+c)Q=a^{2}(b+c+d)+b^{2}(c+d+a)+c^{2}(d+a+b)+d^{2}(a+b+c)

Since
Q=a[a(b+c+d)]+b[b(c+d+a)]+c[c(d+a+b)]+d[d(a+b+c)]=a[a(S−a)]+b[b(S−b)]+c[c(S−c)]+d[d(S−d)]=a[−(a−S2)2+S24]+b[−(b−S2)2+S24]+c[−(c−S2)2+S24]+d[−(d−S2)2+S24]⩽S24(a+b+c+d)=S34\begin{array}{l} Q=a[a(b+c+d)]+b[b(c+d+a)]+ \\ c[c(d+a+b)]+d[d(a+b+c)]= \\ a[a(S-a)]+b[b(S-b)]+c[c(S-c)]+d[d(S-d)]= \\ a\left[-\left(a-\frac{S}{2}\right)^{2}+\frac{S^{2}}{4}\right]+b\left[-\left(b-\frac{S}{2}\right)^{2}+\frac{S^{2}}{4}\right]+ \\ c\left[-\left(c-\frac{S}{2}\right)^{2}+\frac{S^{2}}{4}\right]+d\left[-\left(d-\frac{S}{2}\right)^{2}+\frac{S^{2}}{4}\right] \leqslant \\ \frac{S^{2}}{4}(a+b+c+d)=\frac{S^{3}}{4} \end{array}

Thus, S3=A+6B+3Q⩽A+6B+3S34S^{3}=A+6 B+3 Q \leqslant A+6 B+\frac{3 S^{3}}{4}, which means S3⩽4A+24BS^{3} \leqslant 4 A+24 B.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.