Prove that: $\sum\limits_{cyc} \frac{b+c}{a^{2}+bc}\leq \sum\limits_{cyc} \frac{1}{a}$

Question

If $a$, $b$ and $c$ are positive then $$\sum\limits_{cyc} \frac{b+c}{a^{2}+bc}\leq \sum\limits_{cyc} \frac{1}{a}$$ The things I have done so far: $$\frac{b+c}{a^2+bc}\leq \frac{b+c}{2a\sqrt{bc}}=\frac{b}{2a\sqrt{bc}}+\frac{c}{2a\sqrt{bc}}\leq \sqrt{2\left ( \frac{b^2}{4a^2bc}+\frac{c^2}{4a^2bc} \right )}= \sqrt{2\left ( \frac{b^2+c^2}{4a^2bc}\right )}$$ I just make it looks more difficult!

Answer 1

Let $a\geq b\geq c$.

Thus, $$\sum_{cyc}\frac{1}{a}-\sum_{cyc}\frac{b+c}{a^2+bc}=\sum_{cyc}\frac{(a-b)(a-c)}{a^3+abc}\geq$$ $$\geq\frac{(b-a)(b-c)}{b^3+abc}+\frac{(c-a)(c-b)}{c^3+abc}=(b-c)\left(\frac{a-c}{c^3+abc}-\frac{a-b}{b^3+abc}\right)\geq0.$$

Answer 2

By expanding the statement, the inequality becomes $$a^4b^4+b^4c^4+c^4a^4 \geq a^4b^2c^2+b^4c^2a^2+c^4a^2b^2$$ This inequality is same as the popular $$p^2+q^2+r^2\geq pq + qr + rp$$ In case you do not know how to solve that, $$p^2+q^2+r^2-pq - qr - rp = \frac{1}{2}(p-q)^2+\frac{1}{2}(q-r)^2+\frac{1}{2}(r-p)^2\geq 0$$

side note: btw, this works for your original question as well. There was a constraint $0<a,b,c\leq1$. Assume both sides are multiplied by a non-negative constant $\dfrac{1}{k}$, then left terms become of the form $\dfrac{kb+kc}{k^2a^2+kbkc}$ and the right terms become $\dfrac{1}{ka}$. That indicates if $a, b, c \in (0,1]$ satisfy the inequality, so are $ka, kb, kc$ for any non-negative $k$. In other words, the range constraint does not matter. New constraint would be $a,b,c>0$.

The more you know :)