If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$.

Solution 1:

With the substitutions $a = \frac{x}{y}, \ b = \frac{y}{z}$ and $c = \frac{z}{x}$, it suffices to prove that $$\sum_{\mathrm{cyc}} \sqrt{\frac{zx}{4zx + 2y^2 + 3yz}} \le 1.$$ It suffices to prove that (The desired result follows by summing cyclically.) $$\sqrt{\frac{zx}{4zx + 2y^2 + 3yz}} \le \frac{9x^2+9z^2 + 16xy + 4yz + 34zx}{18x^2+18y^2+18z^2+54xy+54yz+54zx}.$$ Squaring both sides, it suffices to prove that $f(x, y, z) \ge 0$ where \begin{align} f(x,y,z) &= 162 x^4 y^2-549 x^4 y z+504 x^4 z^2+576 x^3 y^3-452 x^3 y^2 z-1300 x^3 y z^2+1708 x^3 z^3+512 x^2 y^4\nonumber\\ &\quad +1144 x^2 y^3 z-1148 x^2 y^2 z^2-1582 x^2 y z^3+504 x^2 z^4-68 x y^4 z-440 x y^3 z^2-596 x y^2 z^3\nonumber\\ &\quad+180 x y z^4+32 y^4 z^2+192 y^3 z^3+378 y^2 z^4+243 y z^5. \end{align} We use the Buffalo Way. There are three possible cases:

1) $z = \min(x,y,z)$: Let $y = z + s, \ x = z+ t; \ s, t\ge 0$. We have $$f(z+t, z+s, z) = a_4z^4 + a_3z^3 + a_2z^2 + a_1z + a_0$$ where \begin{align} a_4 &= 5616 s^2-1728 s t+1728 t^2, \\ a_3 &= 3376 s^3+12864 s^2 t-1176 s t^2+1000 t^3, \\ a_2 &= 476 s^4+7400 s^3 t+10156 s^2 t^2-1376 s t^3+117 t^4, \\ a_1 &= 956 s^4 t+4920 s^3 t^2+1924 s^2 t^3-225 s t^4, \\ a_0 &= 512 s^4 t^2+576 s^3 t^3+162 s^2 t^4. \end{align} It is easy to prove that $a_4\ge 0, \ a_3\ge 0, \ a_2 \ge 0, \ a_0 \ge 0$ and $4a_2a_0 \ge a_1^2$. Thus, $f(z+t, z+s, z) \ge 0$.

2) $y = \min(x,y,z)$: Let $z = y+s, \ x = y+t; \ s, t\ge 0$. We have \begin{align} f(y+t, y, y+s) &= (5616 s^2-1728 s t+1728 t^2) y^4+(6400 s^3+6600 s^2 t+5088 s t^2+1000 t^3) y^3\nonumber\\ &\quad +(2277 s^4+6116 s^3 t+11626 s^2 t^2+3908 s t^3+117 t^4) y^2\nonumber\\ &\quad +(243 s^5+1188 s^4 t+5558 s^3 t^2+5840 s^2 t^3+459 s t^4) y+504 s^4 t^2+1708 s^3 t^3+504 s^2 t^4. \end{align} Clearly, $f(y+t, y, y+s)\ge 0$.

3) $x = \min(x,y,z)$: Similar.