Prove for all positive reals $a,b,c$, $$\sum_{cyc} \frac{a}{\sqrt{3ab+bc}} \ge \frac 32.$$ I'll wait for comments and will post the solution in 2 days! EDIT - Well now that it's been 3 days (oops i forgot to post), I'll show my solution now. By Holder,$$\left( \sum_{cyc} \frac{a}{\sqrt{3ab+bc}} \right)^2 \left( \sum_{cyc} a(3ab+bc) \right) \ge (a+b+c)^3$$ So it suffices to show$$(a+b+c)^3 \ge \frac 94 \sum_{cyc} a(3ab+bc)$$ Expanding and cancelling terms, we wish to show$$f(a,b,c)=4\sum_{cyc} a^3 + 12\sum_{cyc} b^2a-15\sum_{cyc} a^2b-3abc\ge 0$$ Claim: $f(a,b,c) \le f(a+d, b+d, c+d)$ if $d>0$ Proof:$$f(a+d,b+d,c+d)-f(a,b,c) =d (4\sum_{cyc} 3a^2 + 12\sum_{cyc} (b^2+2ba) - 15\sum_{cyc} (a^2+2ba) - 3(ab+bc+ca)) $$ $$+d^2 (4\sum_{cyc} 3a + 12 \sum_{cyc} (2b+a) - 15\sum_{cyc} (2a+b) - 3(a+b+c)$$ The $d^2, d^3$ stuff cancel. So $f(a+d,b+d,c+d) - f(a,b,c) = \frac{9d}{2} \sum_{cyc} (a-b)^2 > 0$. Let $a=\min\{a,b,c\}$, then $f(a,a+b,a+c) \ge f(0,b,c)$. Let $x=\frac bc$, then
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