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The inequality is as follows:

$a^2+b^2+c^2+3\ge2(a+b+c)$

I was thinking about using the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

But I can't get anywhere. I don't have any idea at the moment.

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    You are writing $a^2+b^2+ c^2-3abc$ is it actual question or typing mistake? – 2017-01-15

2 Answers 2

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In the last form you may rewrite it as:

$$(a-1)^2+(b-1)^2+(c-1)^2\ge 0$$

which is obviously true.

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From given equation we have,

$a^2 + b^2 + c^2 − 2a − 2b − 2c + 3 \ge 0$

$(a^2 - 2a +1) + (b^2 - 2b + 1) + (c^2 − 2c + 1) \ge 0$

$(a−1)^2 + (b−1)^2 + (c−1)^2 \ge 0$