# Inequality of *Problem From The Book* 19.20, but Using Integrals

The problem is here:

Let $$a_1,a_2,\dots,a_n$$ be positive real numbers and let $$S=a_1+a_2+\dots+a_n$$. Prove that $$\frac 1n \sum_{1=1}^n\frac 1{a_i}+\frac{n(n-2)}S\ge\sum_{i\ne j}\frac1{S-a_i+a_j}$$

The chapter of this problem is Solving Elementary Inequality Using Integrals. After I typed the problem, I noted that this problem is already asked here. The answer is using Karamata's inequality, which... does not really fit the title of the chapter. This chapter suggested a technique: using $$\int_0^1 x^{t-1}\mathrm dx=\frac 1t$$ to transform the problems involving the fractions into polynomials. I tried this technique, let $$y_i=x^{a_i-1/n}$$, and we have, before integral as $$\int_0^1 \dots\mathrm dx$$, the inequality should be $$\sum_{i=1}^n y_i^n+n(n-2)y_1y_2\dots y_n\ge y_1y_2\dots y_n\sum_{i\ne j}\frac{y_i}{y_j}$$ and I don't know how to continue from here... or, is there another integration method other than the techneque above? Many thanks!

• The resulting inequality is true. Commented Feb 15, 2022 at 9:57
• @RiverLi Thank you for your comment. Would you like to write an answer to this? I believe it is true, but I cannot prove it... Commented Feb 15, 2022 at 13:52

Fact 1: Let $$y_i > 0, \forall i$$. Then $$\sum_{i=1}^n y_i^n + n(n - 1)\prod_{i=1}^n y_i \ge \prod_{i=1}^n y_i \sum_{i=1}^n y_i \sum_{i=1}^n \frac{1}{y_i}. \tag{1}$$ (The proof is given [1], Ch. 5, page 249.)

From Fact 1, letting $$y_i = x^{a_i - 1/n}, \forall i$$, we integrate $$\int_0^1 \cdots \mathrm{d}x$$ both sides of (1) to get the desired result.

Remarks: For $$n = 3$$, (1) is just 3 degree Schur.
For $$n = 4$$, (1) is $$x^4 + y^4 + z^4 + w^4 + 12xyzw \ge (x + y + z + w)(xyz + yzw + zwx + wxy).$$ (I came to know this inequality in AoPS, known as one of Vasc's inequalities.)

Reference

[1] Vasile Cirtoaje, "Algebraic Inequalities-Old and New Methods," 2006.

• Thank you very much for your answer, but I am really sorry that currently I don't have access to the reference source... Would you like to present the proof here? I am really sorry for making you doing me my favor again... Commented Feb 15, 2022 at 15:10