Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.9 AIME, harder Prove it

15. Let $a, b, c \in \mathbf{R}^{+}$ , when $a^{2-}+b^{2}+c^{2}+a b c=4$ , prove: $a+b+c \leqslant 3$ .

Solution

15. Since $a^{2}+b^{2}+c^{2}+a b c=(a+b+c)^{2}-2(a b+b c+c a)+a b c=(a+$ $b+c)^{2}-2(a-2)(b-2)(c-2)-4(a+b+c)+8=4$ , therefore
$(a+b+c)^{2}-2(a-2)(b-2)(c-2)-4(a+b+c)+4=0$

Also, since $a, b, c$ are positive numbers, so $a^{2}>0$ , thus, $0<x \leqslant 3$ . That is, $a+b+c \leqslant 3$ .

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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.