# Is a (short and) beautiful proof for symmetric inequality must exist always?

There are several and almost similar inequalities in MSE that some of them can be proved in long page. some of these questions listed below:

and so on. One cane pose many many similar question in this way: Let $$f(x)$$ be a continuous (and maybe with a special property) then prove that $$\sum_{x\in\{a,b,c\}}f(x)\leq 3f(1)$$ whenever $$abc=1$$. or one can generalize this for arbitrary number of variables: $$\sum_{cyc}f(x)\leq nf(1)$$ whenever $$\prod_{i=1}^n x_i=1$$.

My argument based on what I read in Problem-Solving Through Problems by Loren C. Larson that

principle of insufficient reason, which can be stated briefly as follows: "Where there is no sufficient reason to distinguish, there can be no distinction."

So my question is

Is a (short and) beautiful proof for similar inequalities must exist always as the OPs want for desired answer?

It not always exists, but very very wanted that we can find a nice solution.

I'll give one example.

In 1988 Walther Janous proposed the following problem (Crux 1366).

Let $$a$$, $$b$$ and $$c$$ be positive numbers. Prove that: $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt2}.$$

This problem was an open issue until 2005 before Peter Scholze came and found the following.

We need to prove that: $$\sum_{cyc}\frac{a^2}{\sqrt{a^2+b^2}}\geq\frac{a+b+c}{\sqrt2}$$ or $$\sum_{cyc}\left(\frac{a^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{(a^2+b^2)(b^2+c^2)}}\right)\geq\frac{1}{2}\sum_{cyc}(a^2+2bc),$$ which is true by Rearrangement: $$\sum_{cyc}\left(\frac{a^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{(a^2+b^2)(b^2+c^2)}}\right)-\frac{1}{2}\sum_{cyc}(a^2+2bc)=$$ $$=\sum_{cyc}\left(\frac{1}{2}\frac{a^4+b^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{a^2+b^2}}\cdot\frac{1}{\sqrt{b^2+c^2}}\right)-\frac{1}{2}\sum_{cyc}(a^2+2bc)\geq$$

$$\geq\sum_{cyc}\left(\frac{1}{2}\frac{a^4+b^4}{a^2+b^2}+\frac{2a^2b^2}{\sqrt{a^2+b^2}}\cdot\frac{1}{\sqrt{a^2+b^2}}\right)-\frac{1}{2}\sum_{cyc}(a^2+2bc)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^4}{a^2+b^2}\geq0.$$

If there is so beautiful solution for so hard problem, why we can not try to find something for another problems?

• With computer, the problem can be solved by the Buffalo Way. But to pursuit solutions by hand, Peter Scholze's solution is excellent. – River Li Aug 18 at 2:08
• @River Li In 2005 we had no WolframAlpha and to use BW in 2005 for polynomials with big degree was just impossible. – Michael Rozenberg Aug 18 at 4:19
• Although the solution is pretty short but the openness of problem maybe refer to this point that this problem is not so important so, no attempt to find a solution. – C.F.G Aug 18 at 5:11