Maths Olympiad Prep

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Algebra Difficulty 5.2 AIME, harder Prove it Romania

a. Show that x4−x3−x+1≥0x^4 - x^3 - x + 1 \ge 0, for all real numbers xx.

b. Find all real numbers x1,x2x_1, x_2 and x3x_3 given that x1+x2+x3=3x_1 + x_2 + x_3 = 3 and x13+x23+x33=x14+x24+x34x_1^3 + x_2^3 + x_3^3 = x_1^4 + x_2^4 + x_3^4.

Solution

a. Write x4−x3−x+1=(x−1)(x3−1)=(x−1)2(x2+x+1)x^4 - x^3 - x + 1 = (x-1)(x^3 - 1) = (x-1)^2(x^2 + x + 1) and notice that x2+x+1>0x^2 + x + 1 > 0 for all x∈Rx \in \mathbb{R} to get the claim.

b. Notice that ∑k=13(xk4−xk3−xk+1)=0\sum_{k=1}^{3} (x_k^4 - x_k^3 - x_k + 1) = 0 and use (a) to derive that x14−x13−x1+1=0x_1^4 - x_1^3 - x_1 + 1 = 0, k=1,2,3k = 1, 2, 3. It follows that x1=x2=x3=1x_1 = x_2 = x_3 = 1.

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