Olympiad Maths Prep

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Problem 933

AIME late
Algebra Difficulty 5.8 Prove it

13. Given that a,b,ca, b, c are positive real numbers, a2+b2+c2+abc=4a^{2}+b^{2}+c^{2}+a b c=4, prove: a+b+c≤3a+b+c \leq 3.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

Proof: Without loss of generality, let (b−1)(c−1)≥0(b-1)(c-1) \geq 0, then bc≥b+c−1b c \geq b+c-1.
4−a2=b2+c2+abc≥2bc+abc=bc(2+a),4-a^{2}=b^{2}+c^{2}+a b c \geq 2 b c+a b c=b c(2+a),

i.e., 2−a≥bc≥b+c−12-a \geq b c \geq b+c-1, so a+b+c≤3a+b+c \leq 3.

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