Question: Show that for real $u,x,y,z>0$, $$f(u,x,y,z)=\frac{1}{x^3}+\frac{2z^4}{y^6}+3u^3z^2+\frac{5x^2y^4z^2}{u}$$ has no minimum.
I found the Hessian matrix of $f$ is very complicated, it's hard to show the semi-positive definite property.
My guess is to use some inequality to obtain a constant lower bound of $f$, then verify that the equation can not be reached.
However I cannot find the proper inequality to use.
Any help will be appreciated!
It is obvious that the quantity must be positive, because each individual term in the sum is strictly positive.
Now consider $$f(1,q^{-1/3},q^{1/6},q^{1/2})=q+2q+3q+5q=11q$$ You can obtain any positive value $\epsilon>0$ simply by choosing $$q=\epsilon/11, \quad u=1, \quad x=q^{-1/3}, \quad y=q^{1/6}, \quad z=q^{1/2}.$$ Thus $f$ can be arbitrarily close to zero, but not zero itself, because $q\rightarrow 0$ means that $x\rightarrow +\infty$ and $y,z\rightarrow 0$. So we have $\inf_{u,x,y,z} f(u,x,y,z) =0$, but there is no tuple $(u,x,y,z)$ that attains this infimum.