I have been ponder around a difficult Vietnamese University entrance exam of the year 2003
that is to find the minimum of $\sqrt {{x^2} + \frac{1}{{{x^2}}}} + \sqrt {{y^2} + \frac{1}{{{y^2}}}} + \sqrt {{z^2} + \frac{1}{{{z^2}}}} $ given that $x + y + z \le 1$ ? and x,y,z are all greater of equal to zero
The answer for this problem $\sqrt {82}$, is there anyway we could find the minimum of this problem from the convex optimization point of view because the term $\sqrt {{x^2} + \frac{1}{{{x^2}}}}$ is convex on the interval [0,1]
Also is there any underlying hidden convex structure in this problem (least square, second order cone , quadratic...) ?
Thank you very much for your enthusiasm !
As you point out, the function is convex on the standard simplex $\Delta^2$. Therefore, for all $(x,y,z) \in \Delta^2$, the function value at the average is less than or equal to the average of the function values: $$f(m,m,m) \leq (f(x,y,z) + f(x,z,y) + f(y,x,z) + f(y,z,x) + f(z,x,y) + f(z,y,x))/6,$$ with $m=(x+y+z)/3$. Due to symmetry: $$f(x,y,z) = f(x,z,y) = f(y,x,z) = f(y,z,x) = f(z,x,y) = f(z,y,x),$$ and therefore: $f(m,m,m) \leq f(x,y,z)$ for all $(x,y,z) \in \Delta^2$. The minimum is therefore attained at $x=y=z=1/3$.