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Problem 1046

AMC 12 late, AIME early
Algebra Difficulty 4.9 Prove it Taiwan IMO Selection Camp · Taiwan

Let real numbers a,b,c,da, b, c, d satisfy a+b+c+d=6a + b + c + d = 6 and a2+b2+c2+d2=12a^2 + b^2 + c^2 + d^2 = 12. Prove that
36≤4(a3+b3+c3+d3)−(a4+b4+c4+d4)≤48. 36 \le 4(a^3 + b^3 + c^3 + d^3) - (a^4 + b^4 + c^4 + d^4) \le 48.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Observe that
4(a3+b3+c3+d3)−(a4+b4+c4+d4)=−((a−1)4+(b−1)4+(c−1)4+(d−1)4)+6(a2+b2+c2+d2)−4(a+b+c+d)+4=−((a−1)4+(b−1)4+(c−1)4+(d−1)4)+52. \begin{aligned} & 4(a^{3} + b^{3} + c^{3} + d^{3}) - (a^{4} + b^{4} + c^{4} + d^{4}) \\ &= -((a - 1)^{4} + (b - 1)^{4} + (c - 1)^{4} + (d - 1)^{4}) \\ & \quad +6(a^{2} + b^{2} + c^{2} + d^{2}) - 4(a + b + c + d) + 4 \\ &= -((a - 1)^{4} + (b - 1)^{4} + (c - 1)^{4} + (d - 1)^{4}) + 52. \end{aligned}
Let x=a−1,y=b−1,z=c−1,t=d−1x = a - 1, y = b - 1, z = c - 1, t = d - 1, then it suffices to prove:
under the condition
x2+y2+z2+t2=4(1) x^2 + y^2 + z^2 + t^2 = 4 \qquad (1)
the following inequality holds
16≥x4+y4+z4+t4≥4. 16 \geq x^4 + y^4 + z^4 + t^4 \geq 4.
By the power mean inequality we get
x4+y4+z4+t4≥(x2+y2+z2+t2)24=4(by (1)). x^{4} + y^{4} + z^{4} + t^{4} \geq \frac{(x^{2} + y^{2} + z^{2} + t^{2})^{2}}{4} = 4 \quad (\text{by (1)}).
Next,
(x2+y2+z2+t2)2=(x4+y4+z4+t4)+q, (x^2 + y^2 + z^2 + t^2)^2 = (x^4 + y^4 + z^4 + t^4) + q,
where qq is a nonnegative real number, so
x4+y4+z4+t4≤(x2+y2+z2+t2)2=16. x^{4} + y^{4} + z^{4} + t^{4} \leq (x^{2} + y^{2} + z^{2} + t^{2})^{2} = 16.

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