<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] 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.
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