Let $a$, $b$, $c\ge0$, $abc=1$, prove that \[\sqrt{\frac{a^3+a^2+1}{a^2+a+1}}+\sqrt{\frac{b^3+b^2+1}{b^2+b+1}}+\sqrt{\frac{c^3+c^2+1}{c^2+c+1}}\ge3.\]
The following inequality fails: \[\sqrt{\frac{a^3+a^2+1}{a^2+a+1}}\ge\frac{a+2}3,\] because $o\left(\sqrt{\frac{a^3+a^2+1}{a^2+a+1}}\right)=o\left(\sqrt a\right)<o(a)=o\left(\frac{a+2}3\right)$.
The correct method(s) can be seen in the answer below.
Hint:
It is enough to show for positive $x$ $$\sqrt{\frac{x^3+x^2+1}{x^2+x+1}} \geqslant 1+\frac13\log(x)$$
Another way, we make a different estimate to avoid the need for differentiating logs etc.. Note: $$\frac{x^3+x^2+1}{x^2+x+1}-\frac{2x+1}3 = \frac{(x-1)^2(x+2)}{3(x^2+x+1)}\geqslant 0$$ $$\implies \sqrt{\frac{x^3+x^2+1}{x^2+x+1}}\geqslant \sqrt{\frac{2x+1}3}$$ $$\implies \sum_{x\in\{a, b, c\}} \sqrt{\frac{x^3+x^2+1}{x^2+x+1}}\geqslant \sum_{x\in\{a, b, c\}}\sqrt{\frac{2x+1}3} \\ \geqslant \frac3{\sqrt3}\;\sqrt[6]{(2a+1)(2b+1)(2c+1)} \\ \geqslant \sqrt3 \;\sqrt[6]{(2+1)^3}=3$$
Where we have used AM-GM and Holder's inequalities in the last two steps...