<s>[INST] Prove inequality $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≤ 1.5$ Prove $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≥ 1.5$ with $a + b + c = 3   $ and $a,b,c > 0$
The correct question is $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} ≥ 1.5$ (I have proved it)

Can prove $\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq 1.5$ now:
$\frac{a}{b^2 + 1} = a-\frac{ab^2}{b^2+1}\geq a-\frac{ab}{2}$
Do it the same for $\frac{b}{c^2 + 1}, \frac{c}{a^2 + 1}$, then sum...
$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq a+b+c - \frac{ab+bc+ac}{2} = 3 - \frac{ab+bc+ac}{2}$.
In different side: $(a+b+c)^2 \geq 3(ab+bc+ca) => ab+bc+ac \leq3$.
So:
$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq 3 - 1,5 = 1,5$
The following inequality is true already.

Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1}+\frac{63abc}{2(a^2+b^2+c^2+6(ab+ac+bc))}\leq3.$$

Also, there is the following very nice inequality.

Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that:
$$\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1}+\frac{ab+ac+bc}{2}\geq3.$$

 [/INST] The inequality does not hold! For $a = 2, b = c = 1/2$, we have
$$
\frac{a}{b^2 + 1} + \frac{b}{c^2 + 1} + \frac{c}{a^2 + 1} \geq \frac{a}{b^2 + 1} = \frac 8 5 > \frac 3 2
$$
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