<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] 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.
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