Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$

Suppose $a \geqslant b \geqslant c,$ we have $$\frac{a^2}{a^2+bc} < 1,$$ and $$\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} \leqslant \frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+b^2} = 1.$$ Therefore $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} < 2 .$$

We can rewrite the LHS :

$$LHS = \frac1{1+ \frac{bc}{a^2}} + \frac1{1+\frac{ac}{b^2}} + \frac1{1+\frac{ab}{c^2}}$$

Let $u = \frac{b}{a}$, $v = \frac{a}{c}$, $w = \frac{c}{b}$. Then :

$$LHS = \frac{v}{u+v} + \frac{u}{w+u} + \frac{w}{w+v}$$

We suppose that $u \geq v$ and $u \geq w$ (other cases are similar).

If $w > v$ : $$LHS < \frac{v}{u+v} + \frac{u}{v+u} + \frac{w}{w} = 2$$

If $w \leq v$ :

$$LHS < \frac{v}{v+v} + \frac{u}{u} + \frac{w}{w+w} = 2$$

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Taking $u = n^2$, $w = n$, $v = 1$ :

$$LHS = \frac1{1+n}+\frac{n^2}{n+n^2}+\frac{n}{n+1} \to 2 \quad [n \to \infty]$$

Thus $2$ can't be improved.

let $p=\frac{a^2}{a^2+bc},q=\frac{b^2}{b^2+ac},r=\frac{c^2}{c^2+ab}$ then easy to see $p,q,r\in (0,1)$ Now notice that $$\frac{(1-p)(1-q)(1-r)}{pqr}=1$$ so $$p+q+r=1+pq+qr+rp-2pqr$$ It remains to prove $$pq+qr+rp-2pqr\le 1$$ $$\iff \underbrace{p(1-q)(r-1)}_{\le 0}+\underbrace{q(1-r)(p-1)}_{\le 0}+\underbrace{(1-p)(q-1)}_{\le 0}\le 0$$ which is obvious as each term is $\le 0$ as shown