Prove that$$\frac{1}{\sqrt{14a^2+b^2+c^2}}+\frac{1}{\sqrt{14b^2+a^2+c^2}}+\frac{1}{\sqrt{14c^2+b^2+a^2}}\ge \frac{9}{4(a+b+c)},$$holds for all $a,b,c>0.$
I tried to use C-S $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge \frac{9}{x+y+z}$ and we'll prove that $$\sqrt{14a^2+b^2+c^2}+\sqrt{14b^2+a^2+c^2}+\sqrt{14c^2+b^2+a^2}\le 4(a+b+c)$$ The last inequality is totally wrong.
Hope to see some better ideas. Thank you.
By Holder $$\left(\sum_{cyc}\frac{1}{\sqrt{14a^2+b^2+c^2}}\right)^2\sum_{cyc}(14a^2+b^2+c^2)(4a^2+15b^2+15c^2+48bc+10ab+10ac)^3\geq$$ $$\geq\left(\sum_{cyc}(4a^2+15b^2+15c^2+48bc+10ab+10ac)\right)^3=34^3(a+b+c)^6$$ and it's enough to prove that: $$34^3\cdot16(a+b+c)^8\geq81\sum_{cyc}(14a^2+b^2+c^2)(4a^2+15b^2+15c^2+48bc+10ab+10ac)^3,$$ which is obvious by BW.
We can get a factor $(4a^2+15b^2+15c^2+48bc+10ab+10ac)^3$ by the following reasoning.
For non-negatives $k$, $m$ and $n$ by Holder $$\left(\sum_{cyc}\frac{1}{\sqrt{14a^2+b^2+c^2}}\right)^2\sum_{cyc}(14a^2+b^2+c^2)(ka^2+b^2+c^2+mbc+nab+nac)^3\geq$$ $$\geq\left(\sum_{cyc}(ka^2+b^2+c^2+mbc+nab+nac)\right)^3.$$
The equality occurs for $$\sqrt{14a^2+b^2+c^2}(ka^2+b^2+c^2+mbc+nab+nac)=$$ $$=\sqrt{14b^2+c^2+a^2}(kb^2+c^2+a^2+mca+nab+nbc)=$$ $$=\sqrt{14c^2+a^2+b^2}(kc^2+a^2+b^2+mab+nbc+nac).$$
We know that there are troubles around points $(1,1,0)$ and $(1,0,0)$.
Also, we want to get from $ka^2+b^2+c^2+mbc+nab+nac$ a nice expression, for which we'll assume $m+2n=2(k+2)$.
After solving this system we obtain: $$(k,m,n)\approx\left(\frac{4}{15}\frac{16}{5},\frac{2}{3}\right).$$