Let $f:\mathbb{R}^2\to\mathbb{R}$ be nonnegative and strictly convex in the second argument everywhere (that is, $f(p,\cdot)$ is strictly convex for all $p\in\mathbb{R}$). Moreover, assume $f(p,p)=0$ for all $p$. A primary example of such function is $f(p,q)=(q-p)^2$. If necessary $f$ can be twice-continuously differentiable.

Now, take any $p<q<r$. Is it always the case that the following "reverse triangle inequality'' holds? $$ f(p,q)+f(q,r)<f(p,r) $$ If $f$ is symmetric (i.e., $f(p,q)=f(q,p)$), then I can show this by writing $q=\lambda p + (1-\lambda)r $ for some $\lambda\in(0,1)$. In particular,

\begin{align*} f(p,q)+f(q,r) & =f(p,q)+f(r,q)\\ & =f(p,\lambda p+(1-\lambda)r)+f(r,\lambda p+(1-\lambda)r)\\ & <\lambda f(p,p)+(1-\lambda)f(r,r)+(1-\lambda)f(p,r)+\lambda f(r,p)\\ & =f(p,r) \end{align*}

But, I would like to show this for nonsymmetric $f$, if it is true.

  • $\begingroup$ $f=0$ seems to satisfy your conditions and not satisfy the "strict" triangle inequality. $\endgroup$
    – Kroki
    Feb 26 at 0:06
  • $\begingroup$ @Kroki I added the condition that it is strictly convex $\endgroup$
    – keepfrog
    Feb 26 at 0:57
  • $\begingroup$ @GregMartin Oh shoot, the desired inequality is the other way. I edited the question. $\endgroup$
    – keepfrog
    Feb 26 at 0:59

1 Answer 1


Note that if $f(p,q)$ is a function with the given properties, then so is $h(p,q) = f(p,q)g(p)$ for any positive function $g$ whatsoever. But $h(p,q)+h(q,r)<h(p,r)$ is equivalent to $f(p,q)g(p)+f(q,r)g(q)<f(p,r)g(p)$, or $g(p) \bigl( f(p,q)-f(p,r) \bigr) < g(q) f(q,r)$; and it should be easy to choose a function $g$ for which this is not always true.

For a specific example, let $f(p,q)=(q-p)^2$ and $g(p) = |p|+1$ and set $h(p,q)=f(p,q)g(p)$. Then, choosing $0=p<q<r$, we have $$ h(0,q)+h(q,r)-h(0,r) = q^2\cdot1 + (r-q)^2\cdot(q+1) - r^2\cdot1 = q (r-q) (r-q-2), $$ which is undesirably positive if $q<r<q+2$.


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