Maths Olympiad Prep

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HMIC, 2024

Algebra Difficulty 5.1 AIME, harder Prove it United States

Problem:
Suppose that a,b,ca, b, c, and dd are real numbers such that a+b+c+d=8a+b+c+d=8. Compute the minimum possible value of
20(a2+b2+c2+d2)−∑syma3b, 20\left(a^{2}+b^{2}+c^{2}+d^{2}\right)-\sum_{\text{sym}} a^{3} b,
where the sum is over all 12 symmetric terms.

Solution

Solution:
Observe that
∑syma3b=∑cyca⋅∑cyca3−∑cyca4=8∑cyca3−∑cyca4 \sum_{\mathrm{sym}} a^{3} b=\sum_{\mathrm{cyc}} a \cdot \sum_{\mathrm{cyc}} a^{3}-\sum_{\mathrm{cyc}} a^{4}=8 \sum_{\mathrm{cyc}} a^{3}-\sum_{\mathrm{cyc}} a^{4}
so
20∑cyca2−∑syma3b=∑cyca4−8∑cyca3+20∑cyca2=∑cyc(a4−8a3+20a2)=16(8)−4(4)+∑cyc(a4−8a3+20a2−16a+4)=112+∑cyc(a2−4a+2)2≥112. \begin{aligned} 20 \sum_{\mathrm{cyc}} a^{2}-\sum_{\mathrm{sym}} a^{3} b & =\sum_{\mathrm{cyc}} a^{4}-8 \sum_{\mathrm{cyc}} a^{3}+20 \sum_{\mathrm{cyc}} a^{2} \\ & =\sum_{\mathrm{cyc}}\left(a^{4}-8 a^{3}+20 a^{2}\right) \\ & =16(8)-4(4)+\sum_{\mathrm{cyc}}\left(a^{4}-8 a^{3}+20 a^{2}-16 a+4\right) \\ & =112+\sum_{\mathrm{cyc}}\left(a^{2}-4 a+2\right)^{2} \\ & \geq 112 . \end{aligned}
Equality is achieved when (a,b,c,d)=(2+2,2+2,2−2,2−2)(a, b, c, d)=(2+\sqrt{2}, 2+\sqrt{2}, 2-\sqrt{2}, 2-\sqrt{2}) and permutations. This can be checked by noting that for all x∈{a,b,c,d}x \in\{a, b, c, d\}, we have x2−4x+2=0x^{2}-4 x+2=0, so equality holds in the final step, and for this assignment of variables we have a+b+c+d=8a+b+c+d=8.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.