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 $w \leq v$ :

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

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

From $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2 \iff\\ \frac{a^2+bc-bc}{a^2+bc}+\frac{b^2+ac-ac}{b^2+ac}+\frac{c^2+ab-ab}{c^2+ab}<2 \iff\\ 3-\frac{bc}{a^2+bc}-\frac{ac}{b^2+ac}-\frac{ab}{c^2+ab}<2 \iff\\ \frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}>1$$ and applying Titu's Lemma $$\frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}=\\ \frac{b^2c^2}{a^2bc+b^2c^2}+\frac{a^2c^2}{b^2ac+a^2c^2}+\frac{a^2b^2}{c^2ab+a^2b^2}\geq\\ \frac{(bc+ac+ab)^2}{b^2c^2+a^2c^2+a^2b^2+a^2bc+b^2ac+c^2ab}=\\ \frac{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{2a^2bc+2b^2ac+2c^2ab}}{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{a^2bc+b^2ac+c^2ab}}>1$$

Here's the inevitable 'expand and see if everything works out' solution. $$ \frac{a^2}{a^2 + bc} + \frac{b^2}{b^2 + ac}+\frac{c^2}{c^2 + ab} < 2 \\ \iff a^2 (b^2 + ac)(c^2 + ab) + b^2 (a^2 + bc)(c^2 + ab) + c^2(a^2 + bc)(b^2 + ac) < 2 (a^2 + bc)(b^2 + ac)(c^2 + ab) \\ \iff 3a^2b^2c^2 + 2(a^3b^3 + b^3c^3 + c^3a^3) + a^4 bc + b^4 ac + c^4ab < 2(2a^2b^2c^2 + a^3 b^3 + a^3c^3 + b^3 c^3 + a^4 bc + b^4ac + c^4 ab) $$ It's important when doing these sort of calculations to make sure you didn't miss a term. Here the number of terms on the left (before summing them up) is $12 = 3 \times 2^2$ and the number of terms on the right is $8 = 2^3$ as expected.

This inequality simplifies to $$ a^2b^2c^2 +a^4 bc + b^4ac + c^4 ab > 0 $$ Which is evidently true.