$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$

Question

I was attempting this question: https://www.youtube.com/watch?v=600X-ZGNBbk

Which is the following:

$$\sum_{cyc (a,b,c)}{}\frac{a}{b^2+1} \geq \frac{3}{2}$$

Where $a+b+c = 3$ and $a,b,c > 0 $

The video itself used a solution where they added $3$ and then subtracted $a+b+c$ to rewrite the inequality into something more manageable. Though I understand this solution, it seems arbitrary to me - thinking to add 3 and then subtract the condition is somewhat out of the blue.

What I did was to first try to homogenize the inequality to get rid of the condition. This meant writing $1$ as $\frac{(a+b+c)^2}{9}$ and multiplying the numerator by $\frac{a+b+c}{3}$:

$$\sum_{cyc}^{}\frac{\frac{a(a+b+c)}{3}}{b^2+\frac{(a+b+c)^2}{9}} \geq \frac{3}{2}$$

$$\leftrightarrow $$

$$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$$

I'm fairly certain this inequality is true without the condition of $a+b+c=3$, but I can't see a way to prove it. Multiplying out seems like a lot of trouble; is there an easier way (ideally with Olympiad mathematics methods)?

Answer 1

To prove your inequality (not the original inequality), I would introduce the condition $a + b + c = 1$. Note that \begin{align*} \frac{1}{9b^2+1} &\geq 1 - \frac{3}{2}b\\ \iff 0 &\leq \frac{3}{2}b\left(1 - 3b\right)^2 \end{align*} (this is the Tangent Line Trick). Hence, we have \begin{align*} \sum_{\text{cyc}} \frac{a(a+b+c)}{9b^2+(a+b+c)^2} &= \sum_{\text{cyc}} \frac{a}{9b^2 + 1}\\ &\geq \sum_{\text{cyc}} a\left(1 - \frac{3}{2}b\right)\\ &= 1 - \frac{3}{2}(ab+bc+ca)\\ &\geq 1 - \frac{1}{2}(a+b+c)^2 && \text{by AM-GM}\\ &= \frac{1}{2} \end{align*} and so we're done.

Answer 2

By AM-GM $$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{1+b^2}\geq$$ $$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}\sum_{cyc}ab\geq3-\frac{3}{2}=\frac{3}{2}.$$ Another way.

After full expanding we need to prove that: $$\sum_{cyc}(10a^6+42a^5b+24a^5c-21a^4b^2+105a^4c^2+38a^3b^3)+$$ $$+\sum_{cyc}(84a^4bc-66a^3b^2c+78a^3c^2b-294a^2b^2c^2)\geq0,$$ which is true by AM-GM.

Indeed, $$66\sum_{cyc}a^4c^2=33\sum_{cyc}(a^4c^2+b^4a^2)\geq\sum_{cyc}66a^3b^2c$$ and by Muirhead $$21\sum_{cyc}(a^5b+a^5c)\geq21\sum_{cyc}(a^4b^2+a^4c^2).$$ Thus, it's enough to prove that: $$\sum_{cyc}(10a^6+21a^5b+3a^5c+60a^4c^2+38a^4b^3)+$$ $$+\sum_{cyc}(84a^4bc+78a^3c^2b-294a^2b^2c^2)\geq0,$$ which is AM-GM again.

I hope, now it's clear.