<s>[INST] Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.
and please don't use any $\sum_{cyc}$ because I get confused with it.
 [/INST] 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
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