I'm interested by the following problem :
Let $x,y,z>0$ then we have : $$x+y+z\geq \sqrt{\Big((x^xy^y)^{\frac{1}{x+y}}+(y^yz^z)^{\frac{1}{y+z}}+(z^zx^x)^{\frac{1}{z+x}}\Big)\Big(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\Big)}$$
My try : If we denotes by $f(x,y,z)$ : $$f(x,y,z)=x+y+z-\sqrt{\Big((x^xy^y)^{\frac{1}{x+y}}+(y^yz^z)^{\frac{1}{y+z}}+(z^zx^x)^{\frac{1}{z+x}}\Big)\Big(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\Big)}$$
We have :$$f(x\alpha,y\alpha,z\alpha)=\alpha f(x,y,z)$$
So we can put $x=1$ we get : $$1+y+z\geq \sqrt{\Big((y^y)^{\frac{1}{1+y}}+(y^yz^z)^{\frac{1}{y+z}}+(z^z)^{\frac{1}{1+x}}\Big)\Big(\sqrt{y}+\sqrt{yz}+\sqrt{z}\Big)}$$
Furthermore I have checked the following lemma :
Let $y,z>0$ then we have : $$\sqrt{\Big((y^y)^{\frac{1}{1+y}}+(y^yz^z)^{\frac{1}{y+z}}+(z^z)^{\frac{1}{1+x}}\Big)\Big(\sqrt{y}+\sqrt{yz}+\sqrt{z}\Big)}$$$$\leq$$$$\sqrt{\Big(2(\frac{z+y}{2})^{\frac{\frac{z+y}{2}}{1+\frac{z+y}{2}}}+\frac{z+y}{2}\Big)\Big(2\sqrt{\frac{z+y}{2}}+\frac{z+y}{2}\Big)}$$
So we have to prove that : $$1+y+z\geq\sqrt{\Big(2(\frac{z+y}{2})^{\frac{\frac{z+y}{2}}{1+\frac{z+y}{2}}}+\frac{z+y}{2}\Big)\Big(2\sqrt{\frac{z+y}{2}}+\frac{z+y}{2}\Big)}$$
And you can check that the following function ($z+y=v$): $$f(v)=1+v-\sqrt{\Big(2(\frac{v}{2})^{\frac{\frac{v}{2}}{1+\frac{v}{2}}}+\frac{v}{2}\Big)\Big(2\sqrt{\frac{v}{2}}+\frac{v}{2}\Big)}$$
Is always positive for $v>0$
My question is : How to prove the lemma or have you an alternative proof ?
Thanks in advance for your time .