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.
 A: 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
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ca}+\frac{c^2}{c^2+ab} < 2 .$$
A: 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.
A: 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
A: 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
$$\frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}=\\
\frac{b^2c^2}{a^2bc+b^2c^2}+\frac{a^2c^2}{b^2ac+a^2c^2}+\frac{a^2b^2}{c^2ab+a^2b^2}\geq\\
\frac{(bc+ac+ab)^2}{b^2c^2+a^2c^2+a^2b^2+a^2bc+b^2ac+c^2ab}=\\
\frac{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{2a^2bc+2b^2ac+2c^2ab}}{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{a^2bc+b^2ac+c^2ab}}>1$$
A: 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.
