# Smoothing Inequalities

Can anyone explain to me how the "smoothing argument for inequalities" works?

I know that basically it can be used to prove an inequality $f(a_1,a_2,\cdots,a_n)\geq C$ subject to the constraint $a_1+a_2+\cdots+a_n=k$ (for constants $C,k$) if equality holds when all $a_i$ are equal. I've heard that the basic idea is to push the $a_i$ closer together, showing that this is pushing $f(a_1,a_2,\cdots,a_n)$ to an extrema.

After doing some research, I saw that this can be rigorously defined by showing inequalities of the form $f(a_1,a_2,\cdots,a_n)\geq f(\frac{k}{n},x_1+x_2-\frac{k}{n},\cdots,x_n)$, but I don't really see a pattern here (why do you replace $a_1$ with $\frac{k}{n}$, $a_2$ with $x_1+x_2-\frac{k}{n}$, etc.) Can someone explain this?

Finally, could someone show me how to apply this to an actual inequality problem? For example, how would one solve the following inequality using smoothing (yes, I am aware that there is a quite simple solution using substitutions and AM-GM)?

Let $a_1,\cdots a_n$ be real numbers in the interval $(0,\frac{\pi}{2})$ such that $\sum_{i=0}^n \tan(a_i-\frac{\pi}{4})\geq n-1$. Prove that $\prod_{i=0}^n\tan(a_i)\geq n^{n+1}$

• My edit consisted of replacing "numbners" in the last sentence with "numbers". My most common typo is "sunset" for "subset". Aug 8, 2016 at 18:58

The general idea is that,once you know that a min of some function F exists,it can only occur when all the variables are equal,by showing that if they are not, then a small change in one or two of them will lower the value of F. For the example you give, first some simplification: Let $$b_i= \tan {(a_i- \pi /4)}$$ .Then $$-1 < b_i < 1$$ and $$\tan a_i = (1+b_i)/(1-b_i)$$. Now the logarithm of the product in the example is $$F= \sum ^n_{i=o} \log [(1+b_i)/(1-b_i)].$$ Let us try to minimize this, subject to $$n-1 \le S= \sum ^n_{i=0} b_i$$ and each $$|b_i| < 1$$..... Observe that the second derivative of $$g(x)= \log [(1+x)/(1-x)]$$ is continuous and positive for $$|x| <1$$ which implies that $$2g([x+y]/2) < g(x)+g(y)$$ when $$-1 < x . THEREFORE if $$a_i \ne a_j$$ for some $$i,j$$ then $$b_i \ne b_j$$ so $$2g([b_i+b_j]/2) SO... if we replace each of these unequal $$b_i,b_j$$ with their average then $$F$$ is lowered while $$S$$ is unchanged.So F cannot have a minimum except possibly when all the $$a_i$$ are equal. In this problem,if we do this replacement repeatedly,choosing the largest and smallest $$b_i$$ and $$b_j$$, they will all converge to the average of the original set of $$b_i$$'s, which is in the allowable domain and F is continuous, so the minimum (for a given $$S$$, at least $$n-1$$ ) occurs there.(Note that for a given $$S$$ ,it doesn't matter what the original set is.) For some problems,proving that a minimum exists is easier, for example if the domain is closed and bounded in $$R^n$$ and $$F$$ is continuous, it will.