<s>[INST] If $ab+bc+ca=1$, then $\frac{((a+b)^2+1)}{(c^2+2)}+\frac{((b+c)^2+1)}{(a^2+2)}+\frac{((c+a)^2+1)}{(b^2+2)} \geq 3$ Let $\displaystyle a, b, c> 0, ab+bc+ca=1$. Prove that the following inequality holds:
$$\frac{((a+b)^2+1)}{(c^2+2)}+\frac{((b+c)^2+1)}{(a^2+2)}+\frac{((c+a)^2+1)}{(b^2+2)} \geq 3.$$
I tried using the Cauchy-Schwarz inequality, but can't work it out.
 [/INST] Hint: By CS inequality,$$\sum_{cyc} \frac{(a+b)^2}{(c^2+2)}+\sum_{cyc} \frac1{(c^2+2)}\ge \frac{4(a+b+c)^2+9}{(a^2+b^2+c^2)+6}$$
and then its enough to show 
$$4(a+b+c)^2+9 \ge 3(a^2+b^2+c^2)+18 $$
which should be easy using the constraint given.
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