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!
Best Answer
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.