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Algebra Difficulty 5.3 AIME, harder Prove it

Example 3 Given a,b,c(1,1)a, b, c \in (-1,1).
Prove: abc+2>a+b+cabc + 2 > a + b + c.

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Solution

Proof: Treating aa as a variable and b,cb, c as constants, we denote
f(a)=aix2(a+b+c)=(bc1)a+2bc,a(1,1). \begin{aligned} f(a) & =a i x-2-(a+b+c) \\ & =(b c-1) a+2-b-c, \\ & a \in(-1,1) . \end{aligned}

Now, we only need to prove f(a)>0f(a)>0 based on the properties of a linear function.
b,c(1,1),bcf(1)=1bc+bc=(1b)(1c)>0. \begin{array}{l} \because b, c \in(-1,1), b cf(1)=1-b-c+b c \\ =(1-b)(1-c)>0 . \end{array}

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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.