How to prove or disprove the inequality (from math folklore)
$$\sqrt{22(a^2+b^2+c^2)+5(ab+ac+bc)}\geq\sqrt{4a^2+ab+4b^2}+\sqrt{4b^2+bc+4c^2}+\sqrt{4c^2+ca+4a^2}$$ for nonnegative $a, b,$ and $c$?
My attempt :
Squaring , we get $$ 8\,{a}^{2}+ab+8\,{b}^{2}+2\,\sqrt {4\,{a}^{2}+ab+4\,{b}^{2}}\sqrt {4\, {b}^{2}+bc+4\,{c}^{2}}+2\,\sqrt {4\,{a}^{2}+ab+4\,{b}^{2}}\sqrt {4\,{a }^{2}+ac+4\,{c}^{2}}+bc+8\,{c}^{2}+2\,\sqrt {4\,{b}^{2}+bc+4\,{c}^{2}} \sqrt {4\,{a}^{2}+ac+4\,{c}^{2}}+ac\leq 22\,{a}^{2}+5\,ab+5\,ac+22\,{b }^{2}+5\,bc+22\,{c}^{2}.$$
My googling and search in MSE bring nothing to solve it. Numerical calculations confirm that inequality.
By C-S $\left(\sum\limits_{cyc}\sqrt{4a^2+ab+4b^2}\right)^2\leq\sum\limits_{cyc}(3a+3b+c)\sum\limits_{cyc}\frac{4a^2+ab+4b^2}{3a+3b+c}$.
Thus, it remains to prove that $7(a+b+c)\sum\limits_{cyc}\frac{4a^2+ab+4b^2}{3a+3b+c}\leq\sum\limits_{cyc}(22a^2+5ab)$, which is
$\sum\limits_{cyc}(15a^5+49a^4b+49a^4c-64a^3b^2-64a^3c^2+31a^3bc-16a^2b^2c^2)\geq0$, which is
$15\sum\limits_{cyc}(a^5-a^4b-a^4c+a^3bc)+64\sum\limits_{sym}(a^4b-a^3b^2)+16\sum\limits_{cyc}(a^3bc-a^2b^2c)\geq0$, which is true by Schur and Muirhead.