Let $x,y,z\ge 0$ such $x+y+z=1$, show that $$\sqrt{\dfrac{x(x+1)}{1-x}}+\sqrt{\dfrac{y(y+1)}{1-y}}+\sqrt{\dfrac{z(z+1)}{1-z}}\ge\sqrt{6}$$
This inequality is creat by wangyongxi,and when $x=y=z=\dfrac{1}{3}$ or $x=y=\dfrac{1}{2},z=0$ is $=$.
and we found this $f(x)=\sqrt{\dfrac{x(x+1)}{1-x}},0\le x\le 1$,not Convex function,because we get $$f''(x)=-\dfrac{x^4-4x^3-6x^2-4x+1}{4(x-1)^4\left(-\dfrac{x(x+1)}{x-1}\right)^{3/2}}$$ so when $x<0.2$then $f''(x)<0$ and other $f''(x)>0$
Now I wonder to look such $AM-GM$ or Cauchy inequality to solve it? Thanks
The hint:
We need to prove that $$\sum_{cyc}\sqrt{\frac{x(2x+y+z)}{y+z}}\geq\sqrt{6(x+y+z)}.$$ Indeed, by Holder $$\left(\sqrt{\frac{x(2x+y+z)}{y+z}}\right)^2\sum_{cyc}x^2(2x+y+z)^2(y+z)(x+2y+2z)^3\geq$$ $$\geq\left(\sum_{cyc}x(2x+y+z)(x+2y+2z)\right)^3.$$ Thus, it's enough to prove that: $$\left(\sum_{cyc}x(2x+y+z)(x+2y+2z)\right)^3\geq$$ $$\geq6(x+y+z)\sum_{cyc}x^2(2x+y+z)^2(y+z)(x+2y+2z)^3,$$ which is smooth.
For example, after full expanding we need to prove that $$\sum_{sym}(4x^9+60x^8y+138x^7y^2+37x^6z^3-243x^5y^4+174x^7yz+$$ $$+555x^6y^2z-102x^5y^3z-375x^4y^4z+237x^5y^2z^2-405x^4y^3z^2-80x^3y^3z^3)\geq0.$$ Now, since by Schur $$4\sum_{sym}(x^9-2x^8y+x^7yz)\geq0$$ and $$\sum_{sym}(68x^8y+138x^7y^2+37x^6z^3-243x^5y^4)=$$ $$=\sum_{cyc}xy(x+y)(x-y)^2(68x^4+206x^3y+311x^2y^2+206xy^3+68y^4)\geq0$$ (or for the proof of last inequality we can use AM-GM), it's enough to prove that $$\sum_{sym}(170x^6+555x^5y-102x^4y^2-375x^3y^3+237x^4yz-405x^3y^2z-80x^2y^2z^2)\geq0,$$ which is true by Muirhead.
There is also the following way.
We need to prove that $$\sum_{cyc}\sqrt{\frac{x(2x+y+z)}{(x+y+z)(y+z)}}\geq\sqrt6$$ or $$\sum_{cyc}\frac{2x(2x+y+z)}{2\sqrt{6x(y+z)\cdot(x+y+z)(2x+y+z)}}\geq1$$ and since by AM-GM $$2\sqrt{6x(y+z)\cdot(x+y+z)(2x+y+z)}\leq6x(y+z)+(x+y+z)(2x+y+z),$$ it's enough to prove that $$\sum_{cyc}\frac{2x(2x+y+z)}{6x(y+z)+(x+y+z)(2x+y+z)}\geq1$$ or $$\sum_{cyc}\frac{x}{\frac{6x(y+z)}{2x+y+z}+x+y+z}\geq\frac{1}{2}.$$ Now, by C-S $$\sum_{cyc}\frac{x}{\frac{6x(y+z)}{2x+y+z}+x+y+z}=\sum_{cyc}\frac{x^2}{\frac{6x^2(y+z)}{2x+y+z}+x(x+y+z)}\geq$$ $$\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}\left(\frac{6x^2(y+z)}{2x+y+z}+x(x+y+z)\right)}=\frac{(x+y+z)^2}{6\sum\limits_{cyc}\frac{x^2(y+z)}{2x+y+z}+(x+y+z)^2}.$$ Thus, it remains to prove that $$(x+y+z)^2\geq6\sum\limits_{cyc}\frac{x^2(y+z)}{2x+y+z}$$ or $$\sum_{cyc}\left(x(x+y+z)-\frac{6x^2(y+z)}{2x+y+z}\right)\geq0$$ or $$\sum_{cyc}\frac{x(2x^2+y^2+z^2-3xy-3xz+2yz)}{2x+y+z}\geq0$$ or $$\sum_{cyc}\frac{x(2x^2-3(y+z)x+(y+z)^2)}{2x+y+z}\geq0$$ or $$\sum_{cyc}\frac{x(x-y-z)(2x-y-z)}{2x+y+z}\geq0$$ or $$\sum_{cyc}\frac{x((z-x)(-x+y+z)-(x-y)(-x+y+z))}{2x+y+z}\geq0$$ or $$\sum_{cyc}(x-y)\left(\frac{y(-y+x+z)}{2y+x+z}-\frac{x(-x+y+z)}{2x+y+z}\right)\geq0$$ or $$\sum_{cyc}\frac{(x-y)^2(x^2+4xy+y^2-z^2)}{(2x+y+z)(2y+x+z)}\geq0$$ or $$\sum_{cyc}(x-y)^2(x^2+4xy+y^2-z^2)(2z+x+y)\geq0.$$ Now, let $x\geq y\geq z$.
Thus, $$y^2\sum_{cyc}(x-y)^2(x^2+4xy+y^2-z^2)(2z+x+y)\geq$$ $$\geq y^2\sum_{cyc}(x-y)^2(x^2+y^2-z^2)(2z+x+y)\geq $$ $$\geq y^2(x-z)^2(x^2+z^2-y^2)(2y+x+z)+y^2(y-z)^2(y^2+z^2-x^2)(2x+y+z)\geq$$ $$\geq x^2(y-z)^2(x^2-y^2)(2y+x+z)+y^2(y-z)^2(y^2-x^2)(2x+y+z)=$$ $$=(x^2-y^2)(y-z)^2(x^2(2y+x+z)-y^2(2x+y+z))=$$ $$=(x+y)(x-y)^2(y-z)^2(x^2+y^2+3xy+xz+yz)\geq0.$$ Done!