let $x>0,y>0,z>0$, and $x+y+z=1$,prove that
$$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$
my idea: let $f(x,y)=\dfrac{x^y}{y^x}$
then we consider $$f(x,y)\ge g(x,y)=px+qy+r$$
so $$p=\dfrac{\partial f(x,y)}{\partial x}|_{(x,y)=(1/3,1/3)}=1+\ln{3}$$ $$q=\dfrac{\partial f(x,y)}{\partial y}|_{(x,y)=(1/3,1/3)}=-(1+\ln{3})$$ $$f(1/3,1/3)=g(1/3,1/3)$$ then we have $r=1$
so we have
$$\dfrac{x^y}{y^x}\ge (1+\ln{3})x-(1+\ln{3})y+1$$
my question: How prove that:$$\dfrac{x^y}{y^x}\ge (1+\ln{3})x-(1+\ln{3})y+1$$? a true or flase?
Now we have find this flase. so we can consider other methods. Thank you everyone.
oh,I have consider this methods.
let $$a=\dfrac{x^y}{y^x},b=\dfrac{y^z}{z^y},c=\dfrac{z^x}{x^z}$$
then $$a=y\ln{x}-x\ln{y},b=z\ln{y}-y\ln{z},c=x\ln{z}-z\ln{x}$$ and we have $$az+bx+yc=0$$
so we use this Jensen
$$e^a+e^b+e^c\ge 3e^{\frac{za+xb+yc}{x+y+z}}=3$$
so $$\dfrac{x^y}{y^x}+\dfrac{y^z}{z^y}+\dfrac{z^x}{x^z}\ge 3$$
I belived have other nice methods. Thank you
