Can Jensen be applied to multivariable functions?

Question

I was wondering, suppose we have a symmetric two variable function, in my case: $f(x,y)=\frac{1}{x+y+1}$ with the restriction that $0\le x,y \le 1$, I took the second partial derivative with respect to $x$ which is $\frac{2}{(x+y+1)^3}$, since the function is symmetric I didn't bother taking the partial derivative with respect to $y$, so we see that $f$ is convex in $(0,+\infty)$, now I was wondering, given $0\le a,b,c \le 1$, can I argue that:$$f(a,b)+f(b,c)+f(c,a)\ge 3f(\frac{a+b+c}{3},\frac{a+b+c}{3})$$ Or is this completely non-sense?

Answer 1

You have the function $$g(\rho):={1\over 1+\rho}\qquad(\rho\geq0)\ ,$$ which is certainly convex, and then define $f(x,y):=g(x+y)$. It follows that $$\eqalign{{f(a,b)+f(b,c)+f(c,a)\over3}&={g(a+b)+g(b+c)+g(c+a)\over3}\geq g\left({2a+2b+2c\over3}\right)\cr &=f\left({a+b+c\over3},{a+b+c\over3}\right)\ .\cr}$$Therefore your final statement is correct.

But you can also prove generally that $f$ is convex. Write $(x,y)=:z$ and $\phi(z):=x+y$. Then $f=g\circ\phi$, and $$\eqalign{(1-\lambda)f(z)+\lambda f(z')&=(1-\lambda)g\bigl(\phi(z)\bigr)+\lambda g\bigl(\phi(z')\bigr)\cr &\geq g\bigl((1-\lambda)\phi(z)+\lambda\phi(z')\bigr)=g\bigl(\phi((1-\lambda)z+\lambda z')\bigr)\cr &=f\bigl((1-\lambda)z+\lambda z'\bigr)\ .\cr}$$