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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:

  1. For $abc=1$ prove that $\sum\limits_{cyc}\frac{a}{a^{11}+1}\leq\frac{3}{2}.$
  2. For positive $a$, $b$, $c$ with $abc=1$, show $\sum_{cyc} \left(\frac{a}{a^7+1}\right)^7\leq \sum_{cyc}\left(\frac{a}{a^{11}+1}\right)^7$
  3. Inequality $\frac{x}{x^{10}+1}+\frac{y}{y^{10}+1}+\frac{z}{z^{10}+1}\leq \frac{3}{2}$
  4. Inequality: $(a^3+a+1)(b^3+b+1)(c^3+c+1) \leq 27$
  5. If $abc=1$ so $\sum\limits_{cyc}\frac{a}{a^2+b^2+4}\leq\frac{1}{2}$
  6. For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$
  7. If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$.

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?

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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?

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  • $\begingroup$ With computer, the problem can be solved by the Buffalo Way. But to pursuit solutions by hand, Peter Scholze's solution is excellent. $\endgroup$ – River Li Aug 18 at 2:08
  • $\begingroup$ @River Li In 2005 we had no WolframAlpha and to use BW in 2005 for polynomials with big degree was just impossible. $\endgroup$ – Michael Rozenberg Aug 18 at 4:19
  • $\begingroup$ 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. $\endgroup$ – C.F.G Aug 18 at 5:11

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