Inequality problem involving Cauchy Schwarz inequality If $a+b+c=3$, prove that $\sum \frac{a^2}{b^2-2b+3} \geq 3/2$. How to prove it using the Cauchy-Schwarz inequality? Denote the expression with $P$. What I got was $P$ is greater than or equal to $\frac{9}{\sum a^2-2a-2b-2c+9}$. Is it correct if I found the minimum value of the denominator as $6$ and plugged it in and got the desired result. I mean I got $6$ when equality holds, so I will automatically get the result, wouldn't I?

By C-S $\sum\limits_{cyc}\frac{a^2}{b^2-2b+3}=\sum\limits_{cyc}\frac{a^4}{a^2b^2-2a^2b+3a^2}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2-2a^2b+3a^2)}$.
Hence, it remains to prove that $2(a^2+b^2+c^2)^2\geq3\sum\limits_{cyc}(a^2b^2-2a^2b+3a^2)$, which after homogenization gives
$\sum\limits_{cyc}(a^4+a^2b^2-2a^3c)\geq0$, which is $\sum\limits_{cyc}b^2(a-b)^2\geq0$. Done!
