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We all know the well known upper bound: $$ \min(a,b) \leq a^s b^{1-s}$$ for $$a,b \geq 0, 0 < s < 1$$ I am looking for a lower bound on $\min(a,b)$.

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  • $\begingroup$ What have you tried so far? What are you willing to sacrifice? $\min$ itself is a very good (perfect) lower bound for $\min$, but I expect this is not what you want, so what properties should your lower bound have? $\endgroup$ – flawr Dec 4 '15 at 17:06
  • $\begingroup$ Specificlly I need to lower bound:$$\sum_x \min(p(x),\lambda q(x))$$. where $p,q$ are probability measure. For the upper bound $$\sum_x \min(p(x),\lambda q(x)) \leq \lambda^{1-s} \sum_x p(x)^s q(x)^{1-s} $$ which is close related to renyi divergence $\endgroup$ – nir Dec 4 '15 at 17:17
  • $\begingroup$ What is $p,q,\lambda$ and what values can $x$ take? $\endgroup$ – flawr Dec 4 '15 at 17:19
  • $\begingroup$ also $\lambda > 0$ $\endgroup$ – nir Dec 4 '15 at 17:24
  • $\begingroup$ I do not know any solution, but I am trying some ideas for one, but perhaps I can only post them tomorrow. Seems like an interesting problem=) $\endgroup$ – flawr Dec 4 '15 at 17:46
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You did not specify this, but I assumed you wanted a differentiable lower bound.

If you consider the $f(x,y)=\min(x,y)$ as a function $\mathbb R^2 \to \mathbb R$ and you consider the slice $x+y = const = c$, then my idea was to fit a parabola under that 'triangle' that matches the gradients of $f$ at the zeros. With this idea you end up with following result:

$$\min(x,y) \geq \frac{xy}{x+y} \qquad \forall x\geq0 \forall y\geq 0 \text{ (and $xy\neq 0$)}$$

Proof: WLOG we can assume $y \geq x$. It is obvious that $x(x+y) = x^2+xy \geq xy$. Dividing by $(x+y)$ results in

$$\min(x,y) = x \geq \frac{xy}{x+y}$$

Here a plot of this, the upper surface is $\min$ and the lower is my suggested lower bound.

enter image description here

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  • $\begingroup$ Why the downvotes? $\endgroup$ – flawr Dec 15 '15 at 13:09

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