Factorisation of $b^n-a^n$, and the resulting Lipschitz estimate

The consecutive differences of $c$ factor: for $k<n$, $c_{k+1}-c_k=a^{k+1}{b}^{n-k-1}-a^k{b}^{n-k}=a^k{b}^{n-1-k}a-a^k{b}^{n-1-k}b=a^k{b}^{n-1-k}(a-b)$, using $a^{k+1}=a^{k}a$ and $b^{n-k}=b^{n-1-k}b$.

The endpoints of $c$ are $c_0=a^0{b}^n=b^n$ and $c_n=a^n{b}^0=a^n$.

Now assume $0\le a\le b\le M$; then for every $k<n$, $0\le a^k\le M^k$ and $0\le b^{n-1-k}\le M^{n-1-k}$, so $0\le a^k{b}^{n-1-k}\le M^k{M}^{n-1-k}=M^{n-1}$.

Also $b-a\ge 0$ and $M^{n-1}\ge 0$, and $b^n-a^n\ge 0$ because $0\le a\le b$ gives $a^n\le b^n$.

Telescoping the differences: ${\sum }_{k<n}(c_{k+1}-c_k)=c_n-c_0=a^n-b^n$.

Pulling the constant factor out of the same sum: ${\sum }_{k<n}(c_{k+1}-c_k)={\sum }_{k<n}{a}^k{b}^{n-1-k}(a-b)=(a-b){\sum }_{k<n}{a}^k{b}^{n-1-k}$.

Summing the bound of step 1.3 over $k<n$ and using monotonicity and the constant sum: ${\sum }_{k<n}{a}^k{b}^{n-1-k}\le {\sum }_{k<n}{M}^{n-1}=n{M}^{n-1}$.

Comparing the two evaluations gives $a^n-b^n=(a-b){\sum }_{k<n}{a}^k{b}^{n-1-k}$, and multiplying both sides by $-1$ gives the stated factorisation $b^n-a^n=(b-a){\sum }_{k=0}^{n-1}{a}^k{b}^{n-1-k}$.

Multiplying the inequality of step 2.3 by the nonnegative factor $b-a$ and substituting into the factorisation gives $b^n-a^n=(b-a){\sum }_{k<n}{a}^k{b}^{n-1-k}\le n{M}^{n-1}(b-a)$; the multiplication is the instance of [L5] with $0\le {\sum }_{k<n}{a}^k{b}^{n-1-k}\le n{M}^{n-1}$ and $0\le b-a\le b-a$, the left-hand nonnegativity holding because every term is nonnegative by step 1.3 and finite sums are monotone. Together with $b^n-a^n\ge 0$ this is the Lipschitz estimate.