Just note that $a^4+1 \geq \dfrac{2}{3} \cdot (a^3+a^2+a)$ (it is equivalent to $(a-1)^2(3a^2+4a+3) \geq 0)$, and use AM-GM.

My solution from the exam. $LHS \geq 3\sqrt[3]{\frac{(a^4+1)(b^4+1)(c^4+1)}{(a^3+a^2+a)(b^3+b^2+b)(c^3+c^2+c)}}$ Let's prove that $\frac{a^4+1}{a^3+a^2+a} \geq \frac{2}{3}$ \[3a^4+3 \geq 2a^3+2a^2+2a\]\[(a^2-1)^2+2(a^4-a^3-a+1) \geq 0\]\[(a^2-1)^2+2(a-1)^2(a^2+a+1) \geq0\]The last inequality is true.

Using $x^4+1\ge x^3+x$ (equivalent to $(x-1)^2 (x^2+x+1)\ge 0$), we get that \[ {\rm LHS}\ge 3\sqrt[3]{\prod \frac{a^3+a}{a^3+a^2+a}} = 3\sqrt[3]{\prod \left(1-\frac{a^2}{a^3+a^2+a}\right)}. \]Next, by AM-GM, $a^3+a\ge 2a^2$, so $a^2/(a^3+a^2+a)\le 1/3$. This immediately yields the desired result.

egxa wrote: Prove that for all $a,b,c$ positive real numbers $\dfrac{a^4+1}{b^3+b^2+b}+\dfrac{b^4+1}{c^3+c^2+c}+\dfrac{c^4+1}{a^3+a^2+a} \ge 2$

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Orestis_Lignos wrote: Just note that $a^4+1 \geq \dfrac{2}{3} \cdot (a^3+a^2+a)$ (it is equivalent to $(a-1)^2(3a^2+4a+3) \geq 0)$, and use AM-GM. How did you get, that $a^4+1 \geq \dfrac{2}{3} \cdot (a^3+a^2+a)$ (it is equivalent to $(a-1)^2(3a^2+4a+3) \geq 0)$

Just open the brackets @above