Rearrangement and Chebyshev

We have \(ax+by-ay-bx=(b-a)(y-x)\). Both factors are nonnegative, so the difference is nonnegative.

The difference between the right and left sides is \(ax+cz-az-cx=(c-a)(z-x)\ge0\). Hence the inequality holds.

By the one-swap inequality, \(bz+cx\le bx+cz\), since \(b\le c\) and \(x\le z\). Also \(ay+bx\le ax+by\), since \(a\le b\) and \(x\le y\). Adding gives the result.

Expanding the difference gives \[3\sum ax-\sum a\sum x=(b-a)(y-x)+(c-a)(z-x)+(c-b)(z-y).\] Every term on the right is nonnegative because both sequences are increasing.

Order \(a,b,c\) as \(u_1\le u_2\le u_3\). By rearrangement, \(u_1^2+u_2^2+u_3^2\) is the largest among sums \(\sum u_i u_{\sigma(i)}\). The sum \(ab+bc+ca\) is one of these permutations, so it is not larger than the sum of squares.

If for \(i

After renaming, assume \(x\le y\le z\). Then \(x^2\le y^2\le z^2\). By Chebyshev, \[\frac{x^3+y^3+z^3}{3}\ge\frac{x+y+z}{3}\cdot\frac{x^2+y^2+z^2}{3}.\] Multiplying by \(9\) gives the result.

Order the numbers increasingly. Their squares have the same order. By Chebyshev, \[\frac{\sum a^3}{4}\ge\frac{\sum a}{4}\cdot\frac{\sum a^2}{4}.\] Multiplying by \(16\) gives the inequality.

Let \(u_1\le u_2\le u_3\) be \(a,b,c\) in increasing order. Then \(u_1^2\le u_2^2\le u_3^2\). By rearrangement, the largest sum \(\sum u_i^2u_{\sigma(i)}\) is \(\sum u_i^3\). The right side of the original inequality is one such permutation, so it is not larger.

Order \(a,b,c\) as \(u_1\le u_2\le u_3\). Then \(u_1^3\le u_2^3\le u_3^3\). By rearrangement, the largest product sum \(u_i^3\cdot u_{\sigma(i)}\) is \(u_1^4+u_2^4+u_3^4\). The right side is one of these permutations.

From the cube inequality, \[3\sum a^3\ge\left(\sum a\right)\left(\sum a^2\right).\] Also, \(\sum a^2\ge\frac{(\sum a)^2}{3}\). Therefore \[3\sum a^3\ge \sum a\cdot\frac{(\sum a)^2}{3}=\frac{(\sum a)^3}{3},\] so \(\sum a^3\ge\frac{(\sum a)^3}{9}\).

By Chebyshev, \[\frac{ax+by+cz}{3}\ge\frac{a+b+c}{3}\cdot\frac{x+y+z}{3}=0.\] Hence \(ax+by+cz\ge0\).

The sequences \(a,b,c\) and \(\frac1c,\frac1b,\frac1a\) are both increasing. Therefore \(\frac ac+\frac bb+\frac ca\) is the maximum among all permutations. The sum \(\frac ab+\frac ba+\frac cc\) is one such permutation, so it is not larger.

By rearrangement, the maximum is \(ax+by+cz\), and the minimum is \(az+by+cx\). If some elements among \(a,b,c\) or \(x,y,z\) are equal, several permutations may give the maximum or minimum.

Expanding the right side, the coefficient of \(x\) is \(-b+a-c+a=2a-b-c\), the coefficient of \(y\) is \(b-a-c+b=-a+2b-c\), and the coefficient of \(z\) is \(c-a+c-b=-a-b+2c\). This matches the left side. Since every difference on the right is nonnegative, the left side is nonnegative, which gives Chebyshev.

After ordering, the sequences \(u_i^3\) and \(u_i\) have the same order. By rearrangement, the largest product sum \(u_i^3\cdot u_{\sigma(i)}\) is \(\sum u_i^4\). The cyclic right side is one of the permutations, so it is not larger than \(\sum u_i^4\).

Since the sequences are similarly ordered, Chebyshev gives \[\frac1n\sum_{i=1}^n a_i b_i\ge\left(\frac1n\sum_{i=1}^n a_i\right)\left(\frac1n\sum_{i=1}^n b_i\right)=0.\] Hence \(\sum a_i b_i\ge0\).

After ordering, \(a^2,b^2,c^2\) and \(a^3,b^3,c^3\) increase together. By Chebyshev, \[\frac{a^5+b^5+c^5}{3}\ge\frac{a^2+b^2+c^2}{3}\cdot\frac{a^3+b^3+c^3}{3}.\] Multiplying by \(9\) gives the result.

Reorder the numbers so that \(x_1\le\cdots\le x_n\). Then \(x_1^m\le\cdots\le x_n^m\). Chebyshev gives \[\frac1n\sum x_i^{m+1}\ge\left(\frac1n\sum x_i^m\right)\left(\frac1n\sum x_i\right).\] Multiplying by \(n\) gives the result.

Order \(a,b,c\) as \(u_1\le u_2\le u_3\). Then \(u_i^4\) are ordered in the same way. The maximum of \(\sum u_i^4u_{\sigma(i)}\) is \(\sum u_i^5\). The right side is one of the permutations, so it is not larger.

After ordering \(a,b,c\), we have \(u_1^2\le u_2^2\le u_3^2\) and \(u_1^4\le u_2^4\le u_3^4\). By rearrangement, the largest sum \(u_i^4u_{\sigma(i)}^2\) is \(\sum u_i^6\). The right side corresponds to one permutation, so it is not larger than the left side.

By Chebyshev for \(a^2\) and \(a^3\), \[3\sum a^5\ge(\sum a^2)(\sum a^3).\] Also \(3\sum a^3\ge(\sum a)(\sum a^2)\), and \(\sum a^2\ge\frac{(\sum a)^2}{3}\). Hence \(\sum a^3\ge\frac{(\sum a)^3}{9}\). Substituting this and using \(\sum a^2\ge\frac{(\sum a)^2}{3}\) again, we get \[3\sum a^5\ge\frac{(\sum a)^2}{3}\cdot\frac{(\sum a)^3}{9}=\frac{(\sum a)^5}{27}.\] Therefore \(\sum a^5\ge\frac{(\sum a)^5}{81}\).

By Chebyshev, \[\frac1n\sum_{i=1}^n i\,x_i\ge\left(\frac1n\sum_{i=1}^n i\right)\left(\frac1n\sum_{i=1}^n x_i\right)=0.\] Therefore \(\sum i x_i\ge0\).

Reorder the numbers as \(u_1\le\cdots\le u_n\). Then \(u_1^m\le\cdots\le u_n^m\). By rearrangement, the maximum of \(\sum u_i^m u_{\tau(i)}\) is attained at \(\tau(i)=i\), and equals \(\sum u_i^{m+1}\). The right side of the original inequality corresponds to some permutation \(\tau\), so it is not larger.