I've been trying to prove the following inequality without success. For $a,b,c \in \mathbb{R}$ such that $abc=1$, prove that: $\frac{1}{a^2+a+1}+\frac{1}{b^2+b+1} + \frac{1}{c^2+c+1} \geq 1$
Inequality with constraint
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2@F'OlaYinka Nice observation! I will write up a solution. – 2012-11-12
1 Answers
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.