Let $a,b,c\ge 0: ab+bc+ca+abc=4.$ Prove that
$$\color{black}{\frac{1}{\sqrt{4a^2+4b^2+17ab}}+\frac{1}{\sqrt{4b^2+4c^2+17bc}}+\frac{1}{\sqrt{4c^2+4a^2+17ca}}\ge \frac{3}{5}. }$$ Equality holds at $a=b=c=1; a=b=2,c=0.$
I've tried to use the well-known substitution $a=\dfrac{2x}{y+z};b=\dfrac{2y}{x+z};c=\dfrac{2z}{y+x}$ where $x,y,z\ge 0: xy+yz+zx>0.$
I replace it in the original inequality, $$\sum_{cyc}\frac{1}{\sqrt{4\left(\frac{2x}{y+z}\right)^2+4\left(\frac{2y}{x+z}\right)^2+17\cdot\frac{2x}{y+z}\frac{2y}{x+z}}}\ge \frac{3}{5}.$$$$\sum_{cyc}\frac{1}{\sqrt{\left(\frac{2x}{y+z}\right)^2+\left(\frac{2y}{x+z}\right)^2+17\cdot\frac{x}{y+z}\frac{y}{x+z}}}\ge \frac{6}{5}.$$
$$\sum_{cyc}\frac{(z+x)(z+y)}{\sqrt{17xy(z+x)(z+y)+4(x^2+xz)^2+4(y^2+yz)^2}}\ge \frac{6}{5}.$$ Now, $$17xy(z+x)(z+y)+4(x^2+xz)^2+4(y^2+yz)^2=17xy(z^2+xy+yz+zx)+4(x^4+2x^3z+x^2z^2)^2+4(y^4+2y^3z+y^2z^2)^2$$ $$=4z^2(x^2+y^2)+8z(x^3+y^3)+4(x^4+y^4)+17xyz^2+17xy(xy+yz+zx).$$
From here, I didn't see any bright ideas. Hope you help me.
Bacteria helps!
By Holder $$\left(\sum_{cyc}\frac{1}{\sqrt{4a^2+4b^2+17ab}}\right)^2\sum_{cyc}(4a^2+4b^2+17ab)(3c+4)^3\geq\left(\sum_{cyc}(3c+4)\right)^3$$ and it's enough to prove that: $$75(a+b+c+4)^3\geq\sum_{cyc}(4a^2+4b^2+17ab)(3c+4)^3,$$ which is $f(u)\geq0,$ where $f$ increases(about $uvw$ see here: https://artofproblemsolving.com/community/c6h278791).
Thus, since the condition does not depend of $u$, by $uvw$ it's enough to prove the last inequality for equality case of two variables.
Let $b=a$.
Thus, the condition gives $c=\frac{2-a}{a},$ where $0<a\leq2$ and after this substitution we obtain: $$(a-1)^2(2-a)(216a^5+810a^4+1092a^3+1035a^2+1076a+300)\geq0.$$
Can you obtain now a full solution?
For the proof that $f$ increases use $u\geq v^2$, which is true by C-S and $w^3\leq1$ which is true by AM-GM.