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Let $a,b,c > 0$ such that $a^2 + b^2 + c^2 =3$. Prove that

$$(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$$

My attempt :

Let $\lambda = \prod \limits_{cyc} a^3+a+1$ .

Applying AM-GM on the set $\{(a^3+a+1),(b^3+b+1),(c^3+c+1)\}$ :

$$\lambda^{\frac{1}{3}}\leq\dfrac{\sum \limits_{cyc} a^3+a^2+a}{3} \leq \dfrac{\sum \limits_{cyc} a^3 +2a^2 }{3} = 2+\dfrac{a^3+b^3+c^3}{3}$$

I am struck now and do not how to proceed. The question seems easy but I can not figure out the right strategy. Also, C-S does not seem to help.

Thanks in Advance !

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    I've read this somewhere before (which is why the proof has no context). Beats me where the source was. – 2017-02-14
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    @S.C.B. It is actually a problem from AoPS.... – 2017-02-14
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    Really? I think I saw it somewhere else. – 2017-02-14
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    Here is the [link](https://artofproblemsolving.com/community/c6t321f6h517578_inequality__from_gazeta_matematica) – 2017-02-14
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    So do you understand my answer? – 2017-02-14
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    There are other AoPS questions with the same idea. – 2017-02-14
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    For example, @Nirbhay see [here](https://www.artofproblemsolving.com/community/c6h511786p2873589). – 2017-02-14
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    Why the downvotes ? – 2017-02-16
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    Just try to ignore them. – 2017-02-16

1 Answers 1

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From the expansion of $a^4-4a^3+6a^2-4a+1=(a-1)^{4} \ge 0$, we can simply the equation to get the following: $$4a^3+4a+4 \le a^4+6a^2+5 $$

So we have that $$\prod_{cyc}(a^3+a+1)\leq\frac{1}{4^3}\prod_{cyc}(a^2+1)\prod_{cyc}(a^2+5) \le 27$$ From $\text{AM-GM}$ Inequality. We have the desired result.