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I've been trying to prove the following inequality without success. For $a,b,c \in \mathbb{R}$ such that $abc=1$, prove that: $$\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1} + \frac{1}{c^2+c+1} \geq 1$$

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are $a,b,c$ positive? –  user31280 Nov 12 '12 at 21:31
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If you clear denominators and group terms, the question inequality "simplifies" to $a^2 + b^2 + c^2 \geq ac + bc + ab$. Perhaps someone sees where to go from here. –  Austin Mohr Nov 12 '12 at 21:37
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@AustinMohr the inequality is obvious from there since $(a-b)^2+(b-c)^2+(c-a)^2\ge 0 \iff a^2+b^2+c^2\ge ab+bc+ca$. You should put up your solution for the OP. –  user31280 Nov 12 '12 at 22:19
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@F'OlaYinka Nice observation! I will write up a solution. –  Austin Mohr Nov 12 '12 at 22:22

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up vote 4 down vote accepted

Multiply both sides of the inequality by $(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1)$ to get $$ \begin{align*} &(b^2 + b + 1)(c^2 + c + 1) + (a^2 + a + 1)(c^2 + c + 1) + (a^2 + a + 1)(b^2 + b + 1)\\ \geq &(a^2 + a + 1)(b^2 + b + 1)(c^2 + c + 1). \end{align*} $$ Suffer through the algebra (or enlist Maple's help) to simplify this to $$ a^2 + b^2 + c^2 - ab - ac - bc \geq 0. $$ Note that during the simplification, we use the constraint $abc = 1$ several times.

Now use F'OlaYinka's observation: The inequality above is true if and only if $$ 2(a^2 + b^2 + c^2 - ab - ac - bc) \geq 0. $$ The lefthand side of this expression can be factored as $$ (a-b)^2 + (a-c)^2 + (b-c)^2. $$ Since this is the sum of three positive terms, it is surely greater than or equal to zero, which completes the proof.

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