A cyclic inequality $\sum\limits_{cyc}{\sqrt{3x+\frac{1}{y}}}\geqslant 6$

Given positive real numbers $x,y,z$ satisfying $x+y+z=3$, prove that $$ \sqrt{3x+\frac{1}{y}}+\sqrt{3y+\frac{1}{z}}+\sqrt{3z+\frac{1}{x}}\geqslant 6. $$

Solution 1:

square both sides,we have:

$3(x+y+z)+\sum\limits_{cyc}{\dfrac{1}{x}}+2\sum\limits_{cyc}{\sqrt{3x+\dfrac{1}{y}}\sqrt{3y+\dfrac{1}{z}}}\geqslant 36 \iff \\\sum\limits_{cyc}{\dfrac{1}{x}}+2\sum\limits_{cyc}\left({2\sqrt{xy}+1+\sqrt{\dfrac{x}{z}}} \right) \ge 27$

because $(3x+\dfrac{1}{y})(3y+\dfrac{1}{z})=(x+x+x+\dfrac{1}{y})(\dfrac{1}{z}+y+y+y)\ge \left(\sqrt{\dfrac{x}{z}}+\sqrt{xy}+\sqrt{xy}+1 \right)^2$

$\iff \sum\limits_{cyc}{\dfrac{1}{x}}+4\sum\limits_{cyc}\sqrt{xy} +2\sum\limits_{cyc}\sqrt{\dfrac{x}{z}} \ge 21$

let $a=\sqrt{x},b=\sqrt{y},c=\sqrt{z} \iff a^2+b^2+c^2=3 $

then it becomes:

$\sum\limits_{cyc}{\dfrac{1}{a^2}}+4\sum\limits_{cyc}ab +2\sum\limits_{cyc}\dfrac{b}{a} \ge 21$

WLOG, let $a$ is min of {$a,b,c$}, $b=a+u,c=a+v,u\ge 0,v\ge 0$

both sides ,multiply $a^2+b^2+c^2$,and put $b=a+u,c=a+v$ ,we get:

$6(a^2+(a+u)^2+(a+v)^2)a(a+u)(a+v)(a^2(a+u)+(a+u)^2(a+v)+(a+v)^2a)+(a^2+(a+u)^2+(a+v)^2)^2(a^2(a+u)^2+(a+u)^2(a+v)^2+a^2(a+v)^2)+36(a(a+u)(a+v))^2(a(a+u)+(a+u)(a+v)+(a+v)a)-63(a(a+u)(a+v))^2(a^2+(a+u)^2+(a+v)^2) \ge 0 \iff u^2*v^6+2*a*u*v^6+2*a^2*v^6+6*a*u^2*v^5+18*a^2*u*v^5+18*a^3*v^5+2*u^4*v^4+14*a*u^3*v^4-13*a^2*u^2*v^4-30*a^3*u*v^4+12*a^4*v^4+8*a*u^4*v^3+98*a^2*u^3*v^3+76*a^3*u^2*v^3-44*a^4*u*v^3-2*a^5*v^3+u^6*v^2+12*a*u^5*v^2-a^2*u^4*v^2+100*a^3*u^3*v^2+146*a^4*u^2*v^2+12*a^5*u*v^2+6*a^6*v^2+2*a*u^6*v+24*a^2*u^5*v-12*a^3*u^4*v-14*a^4*u^3*v+30*a^5*u^2*v-6*a^6*u*v+2*a^2*u^6+18*a^3*u^5+12*a^4*u^4-2*a^5*u^3+6*a^6*u^2 \iff \\ \\k_6a^6+k_5a^5+k_4a^4+k_3a^3+k_2a^2+k_1a+k_0 \ge 0$

where :

$k_6=6(u^2-uv+v^2) \ge 0 \\k_5= 2(15u^2v+6uv^2-u^3-v^3)\\k_4=(12u^4-14u^3v+146u^2v^2-44uv^3+12v^4) \ge 0 \\k_3=(18u^5-12u^4v+100u^3v^2+76u^2v^3- 30uv^4+18v^5) \ge 0 \\ k_2= (2u^6+24u^5v-u^4v^2+98u^3v^3-13u^2v^4+18uv^5+2v^6) \ge 0 \\ k_1=uv(12u^4v+14u^2v^3+6uv^4+2v^5) \ge0 \\ k_0=u^2v^2(u^2+v^2)^2 \ge 0$

Edit:

due to $k_5$ is uncertain, now we prove $4k_6*k_4 \ge k_5^2 \iff 4*6(u^2-uv+v^2)*(12u^4-14u^3v+146u^2v^2-44uv^3+12v^4)-(2(15u^2v+6uv^2-u^3-v^3))^2 \ge0 \iff 71*v^6-324*u*v^5+1206*u^2*v^4-1406*u^3*v^3+819*u^4*v^2-126*u^5*v+71*u^6 \ge 0 \iff (71v^6-324uv^5+370u^2v^4)+(734u^2v^4-1406u^3v^3+734u^4v^2)+(85u^4v^2-126u^5v+71u^6) \ge 0$ it is trivial.

that is $k_6a^6+k_5a^5+k_4a^4\ge 0$

it is trivial only when $u=v=0$ the all "=" is hold

QED.

Solution 2:

Remark: @chenbai gave a nice solution using the Buffalo Way (BW).
Here, I give an alternative solution using BW:

By Holder, we have \begin{align} \left(\sum_{\mathrm{cyc}} \sqrt{3x + 1/y}\right)^2 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y} \ge \left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3. \end{align} It suffices to prove that $$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 36 \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + 1/y}.$$ After homogenization, it suffices to prove that, for all $x, y, z > 0$, $$\left(\sum_{\mathrm{cyc}} (6x + y + 3z)\right)^3 \ge 12(x+y+z) \sum_{\mathrm{cyc}} \frac{(6x + y + 3z)^3}{3x + \frac{(x+y+z)^2}{9y}}.$$

The Buffalo Way kills it. We are done.