For $a,b,c>0.$ Prove$:$ $$\sum \dfrac{b+c}{9(a^2+3bc)+4(a+b+c)^2}\geqslant \dfrac{1}{4(a+b+c)}$$
SOS solution$:$ $$\dfrac{1}{8(a+b+c)}\sum{\dfrac { \left( 52\,{a}^{2}+95\,ab-142\,ac+52\,{b}^{2}-142\,bc+103\,{c
}^{2} \right) \left( a-b \right) ^{2}}{ \left( 13\,{a}^{2}+35\,bc+8\,
ab+8\,ac+4\,{b}^{2}+4\,{c}^{2} \right) \left( 13\,{b}^{2}+35\,ac+4\,{
a}^{2}+8\,ab+8\,bc+4\,{c}^{2} \right) }}\geqslant 0,$$
which is clearly true.
I also found a proof by pqr method$:$
Need to prove$:$ $$208\,{p}^{6}-432\,{p}^{4}q-5805\,{p}^{3}r-972\,{p}^{2}{q}^{2}+32724\,p
qr-2187\,{q}^{3}-46656\,{r}^{2} \geqslant 1728(-4{p}^{3}r+{p}^{2}{q}^{2}+18pqr-4{q}^{3}-27{r}^{2})\geqslant 0$$
Note that $$1728(-4{p}^{3}r+{p}^{2}{q}^{2}+18pqr-4{q}^{3}-27{r}^{2})=1728(a-b)^2(b-c)^2(c-a)^2 \geqslant 0.$$
We are only need to prove the left.
It's $$27\,p \left( 41\,{p}^{2}+60\,q \right) r+208\,{p}^{6}-432\,{p}^{4}q-
2700\,{p}^{2}{q}^{2}+4725\,{q}^{3} \geqslant 0.$$
By Schur degree $6$ we have$:$ $$r\geqslant -\frac49\,{p}^{3}+{\frac {11}{9}}\,pq+\frac{1}{9}\sqrt {7\,{p}^{6}-25\,{p}^{4}q+
4\,{p}^{2}{q}^{2}+36\,{q}^{3}}.$$
Let $p=1$ then $q \leqslant \frac{p^2}{3} =\frac{1}{3}$. Hence it's enough to prove$:$
$$-284+201\,q+123\,\sqrt {7-25\,q+4\,{q}^{2}+36\,{q}^{3}}-720\,{q}^{2}+
180\,q\sqrt {7-25\,q+4\,{q}^{2}+36\,{q}^{3}}+4725\,{q}^{3} \geqslant 0.$$
Since $$284-201\,q+720\,{q}^{2}-4725\,{q}^{3} \geqslant 0 \quad \forall \quad 0< q\leqslant \frac{1}{3}$$
So we are only need to prove $$\left( 1-3\,q \right) \left( 7441875\,{q}^{5}-176175\,{q}^{4}+
172665\,{q}^{3}-904113\,{q}^{2}+121644\,q+25247 \right) \geqslant 0,$$
which is true because $0< q\leqslant \frac{1}{3}.$
Done.
Is there any other proof (without Buffalo Way of course)$?$
Best Answer
SOS helps!
We need to prove that: $$\sum_{cyc}\frac{b+c}{13a^2+4b^2+4c^2+8ab+8ac+35bc}\geq\frac{1}{4(a+b+c)}$$ or $$\sum_{cyc}\left(\frac{b+c}{13a^2+4b^2+4c^2+8ab+8ac+35bc}-\frac{1}{12(a+b+c)}\right)\geq0$$ or $$\sum_{cyc}\frac{8b^2+8c^2-13a^2+4ab+4ac-11bc}{13a^2+4b^2+4c^2+8ab+8ac+35bc}\geq0$$ or $$\sum_{cyc}\frac{(c-a)(13a-11b+16c)-(a-b)(13b-11c+16a)}{13a^2+4b^2+4c^2+8ab+8ac+35bc}$$ or $$\sum_{cyc}(a-b)\left(\tfrac{13b-11c+16a}{13b^2+4a^2+4c^2+8ab+8bc+35ac}-\tfrac{13a-11c+16b}{13a^2+4b^2+4c^2+8ab+8ac+35bc}\right)\geq0$$ or $$\sum_{cyc}\tfrac{(a-b)^2(52a^2+95ab+52b^2-142(a+b)c+103c^2)}{(13a^2+4b^2+4c^2+8ab+8ac+35bc)(13b^2+4a^2+4c^2+8ab+8bc+35ac)}\geq0,$$ which is true by C-S and AM-GM: $$52a^2+95ab+52b^2-142(a+b)c+103c^2=$$ $$=\frac{95}{2}(a+b)^2+\frac{9}{2}(a^2+b^2)+103c^2-142(a+b)c\geq$$ $$\geq\frac{95}{2}(a+b)^2+\frac{9}{4}(a+b)^2+103c^2-142(a+b)c=$$ $$=\frac{199}{4}(a+b)^2+103c^2-142(a+b)c\geq\left(\sqrt{199\cdot103}-142\right)(a+b)c\geq0.$$