Five finite proofs of Bernoulli, with rational exponents and the quadratic refinement

The degenerate cases, common to all routes: if $n=0$ both sides are $1$; if $n=1$ both sides are $1+x$; if $x=0$ both sides are $1$, since $1^n=1$ for every natural $n$; so equality holds whenever $n\le 1$ or $x=0$, and it remains to prove the strict inequality $(1+x{)}^n>1+nx$ for $n\ge 2$ and $x\mathrm{\ne }0$ with $x\ge -1$.

Route 1, ordinary induction: the published theorem [L5] states the inequality with exactly the claimed equality clause and its published proof is the ordinary induction, so this route is verified by quoting [L5] as a whole; no duplicate induction is minted.

Route 2, geometric-sum comparison: for $n\ge 2$ and $y\ge 0$, $y\mathrm{\ne }1$, [L6] gives $y^n-1=(y-1){\sum }_{k<n}{y}^k$; if $y>1$ then every term $y^k$ is at least $1$ and the term $k=1$ is strictly larger, so the sum is strictly greater than $n$ by the strict sum comparison of [L1] and $y-1>0$ gives $y^n-1>n(y-1)$, while if $0\le y<1$ every term is at most $1$ and the term $k=1$ is strictly smaller, so the sum is strictly less than $n$ and multiplication by the negative $y-1$ reverses the comparison in [L2] to give again $y^n-1>n(y-1)$; substituting $y=1+x$ gives $(1+x{)}^n>1+nx$ for $n\ge 2$, $x\mathrm{\ne }0$.

Route 3, exact nonnegative remainder: for $n\ge 2$ and $y\ge 0$ one has the identity $y^n-ny+(n-1)=(y-1{)}^2{\sum }_{j=0}^{n-2}(n-1-j)y^j$: indeed [L6] gives $y^n-1=(y-1){\sum }_{k<n}{y}^k$, so $y^n-ny+(n-1)=(y-1)({\sum }_{k<n}{y}^k-n)$, and ${\sum }_{k<n}{y}^k-n={\sum }_{k=1}^{n-1}(y^k-1)={\sum }_{k=1}^{n-1}(y-1){\sum }_{j<k}{y}^j=(y-1){\sum }_{j=0}^{n-2}(n-1-j)y^j$, since each $y^j$ occurs once for each $k>j$ with $k\le n-1$, of which there are $n-1-j$; every coefficient $n-1-j$ is a positive natural and every $y^j\ge 0$, so the last sum is at least its constant term $n-1>0$, and $(y-1{)}^2\ge 0$ with equality only at $y=1$ by [L4]; hence $y^n-ny+(n-1)>0$ for $y\mathrm{\ne }1$, that is $(1+x{)}^n>1+nx$ for $n\ge 2$ and $x\mathrm{\ne }0$, with the identity giving $0=0$ and equality at $x=0$.

Route 4, AM-GM applied to $(y^n,1,\dots ,1)$: for $n\ge 2$ and $y\ge 0$ the $n$ entries $y^n,1,\dots ,1$ are nonnegative, their product is $y^n$, and their geometric mean is $(y^n{)}^{1\mathrm{/}n}=y$ because $y\ge 0$ and $y^n=y^n$; [L7] gives $y\le (y^n+(n-1))\mathrm{/}n$, that is $y^n\ge ny-(n-1)=1+n(y-1)=1+nx$; if $y\mathrm{\ne }1$ the list is not constant, so [L7]'s equality clause makes the inequality strict, and $y=1$ means $x=0$; this proves the integer statement for $n\ge 2$ by a fourth route, with strictness for $x\mathrm{\ne }0$.

Route 5, finite Pascal rows: set $b_{0,0}=1$, $b_{0,k}=0$ for $k\ge 1$, and define $b_{m+1,0}=1$, $b_{m+1,k}=b_{m,k}+b_{m,k-1}$ for $k\ge 1$. This is recursion on $m$ in the set of functions $\mathbb{N}\to \mathbb{N}$, with a successor map sending a row to the displayed new row. Coefficients in field expressions denote their canonical images. Induction gives $b_{m,k}=0$ for $k>m$, since both predecessor entries then vanish, and all coefficients are nonnegative. It also gives $b_{m,0}=b_{m,m}=1$ and $b_{m,1}=m$: the diagonal successor is $0+1$, and the first column increases by $1$. The expansion $(u+v{)}^m={\sum }_{k=0}^m{b}_{m,k}{u}^k{v}^{m-k}$ holds at $m=0$; multiplying the row-$m$ expansion by $u+v$ and shifting the $u$-part by one index gives the coefficient $b_{m,k}+b_{m,k-1}$ at each interior index, and coefficient $1$ at the endpoints, proving the successor expansion. Finally $k{b}_{m,k}=m{b}_{m-1,k-1}$ for $1\le k\le m$, by induction on $m$. At $m=1$ it is $1=1$, and the cases $k=1,m$ follow from the endpoint and first-column values. For $2\le k\le m-1$, the preceding-row identities and Pascal recursion give $k{b}_{m,k}=k{b}_{m-1,k}+k{b}_{m-1,k-1}=(m-1)b_{m-2,k-1}+(m-1)b_{m-2,k-2}+b_{m-1,k-1}=m{b}_{m-1,k-1}$, closing the induction.

Strictness transfer for weighted AM-GM: let $x_0,\dots ,x_{m-1}>0$ and let $w_k\ge 0$ be rationals with ${\sum }_{k<m}{w}_k=1$ as real numbers. The common-denominator and repeated-block construction in the proof of Weighted AM-GM inequality with rational weights, steps 1.1 through 5.1, gives $w_k=n_k\mathrm{/}N$ with $n_k$ natural, $N\ge 1$, and a list of $N$ positive entries, repeating $x_k$ exactly $n_k$ times. Its sum is ${\sum }_{k<m}{n}_k{x}_k$ and its product is ${\prod }_{k<m}{x}_k^{n_k}$. Put $H:={\prod }_{k<m}{x}_k^{w_k}>0$ and $M:={\sum }_{k<m}{w}_k{x}_k>0$. The iterated-power law in [L10], applied to each factor, and the natural product law in [L3], applied inductively, give $H^N={\prod }_{k<m}{x}_k^{n_k}$; the repeated list has mean $M$. AM-GM in [L7] therefore gives $H^N\le M^N$, with equality exactly when the repeated list is constant. By [L3], $H\le M$ with equality under the same condition; in particular the weighted inequality is strict whenever two bases carrying positive weights differ.

Rational Bernoulli forms, inequality directions, for $y>0$ and rational $r$: (a) if $r>1$, the weights $1\mathrm{/}r$ and $1-1\mathrm{/}r$ are rational, nonnegative and sum to $1$, so [L7] applied to the positive bases $y^r$ and $1$ gives $y=(y^r{)}^{1\mathrm{/}r}{1}^{1-1\mathrm{/}r}\le (1\mathrm{/}r)y^r+(1-1\mathrm{/}r)$, and multiplying by the positive $r$ gives $y^r\ge 1+r(y-1)$; (b) if $r<0$, the weights $-r\mathrm{/}(1-r)$ and $1\mathrm{/}(1-r)$ are rational, nonnegative and sum to $1$, and with bases $y$ and $y^r$ the geometric side is $y^{-r\mathrm{/}(1-r)}(y^r{)}^{1\mathrm{/}(1-r)}=y^0=1$ by the addition law of [L10], so $1\le (-ry+y^r)\mathrm{/}(1-r)$ and multiplying by the positive $1-r$ gives $y^r\ge 1+r(y-1)$; (c) if $0<r<1$, the weights $r$ and $1-r$ give $y^r=y^r{1}^{1-r}\le ry+(1-r)=1+r(y-1)$; in every case the powers of $1$ used are $1$ by [L10].

Route 5 for $x>0$: the expansion of step 1.6 at $(u,v)=(x,1)$ gives $(1+x{)}^n={\sum }_{k=0}^n{b}_{n,k}{x}^k=1+nx+{\sum }_{k=2}^n{b}_{n,k}{x}^k$ with $b_{n,0}=1$, $b_{n,1}=n$, all coefficients $\ge 0$ and $b_{n,n}=1$, so $(1+x{)}^n\ge 1+nx+x^n>1+nx$ for $n\ge 2$ and $x>0$; equality at $x\mathrm{\ne }0$, $n\ge 2$ is impossible in this range.

Route 5 for $-1\le x<0$: put $t:=-x\in (0,1]$ and $y:=1-t=1+x\ge 0$; the expansion of step 1.6 gives $1=(t+y{)}^n={\sum }_{k=0}^n{b}_{n,k}{t}^k{y}^{n-k}$ and, using the coefficient identity $k{b}_{n,k}=n{b}_{n-1,k-1}$ and the expansion at $n-1$, also $nt={\sum }_{k=1}^{n}k{b}_{n,k}{t}^k{y}^{n-k}$; subtracting the first identity from the second, $nt-(1-y^n)={\sum }_{k=1}^n(k-1)b_{n,k}{t}^k{y}^{n-k}\ge 0$ with $k=n$ term $(n-1)t^n>0$ for $n\ge 2$, so $nt>1-y^n$ and $y^n>1-nt=1+nx$; this proves the integer statement for the negative range $x\in [-1,0)$ by positive weights, without discarding alternating terms.

Rational equality and strictness: for $y>0$ and rational $r\mathrm{\ne }0,1$, if $x\mathrm{\ne }0$, that is $y\mathrm{\ne }1$, then in each case of step 1.8 the two bases carrying positive weight differ: for $r>1$, $y^r=1$ would force $y=1$ by [L10]; for $0<r<1$ the bases are $y$ and $1$; and for $r<0$, $y^r=y$ would imply $y^{r-1}=1$, hence $y^{1-r}=1$ by taking the inverse through [L10], again forcing $y=1$ since $1-r>0$. Thus the strictness transfer of step 1.7 makes the weighted inequality strict and, after the positive multiplications of step 1.8, the rational Bernoulli inequality is strict as well; if $x=0$, that is $y=1$, every base equals $1$ and both sides equal $1$, and the cases $r=0$ and $r=1$ give $y^0=1=1+0$ and $y^1=y=1+x$ identically, so equality holds exactly for $x=0$ or $r\in \{0,1\}$.

Endpoint $x=-1$: for rational $r>0$ the supplementary clause gives $0^r=0$ by [L10], so the left-hand side is $0$ and the right-hand side is $1-r$; for $r\ge 1$ the assertion $0\ge 1-r$ holds with equality exactly at $r=1$, and for $0<r\le 1$ the reversed assertion $0\le 1-r$ holds with equality exactly at $r=1$; at $r=0$ the integer convention gives $0^0=1=1+0$, an equality, and negative rational exponents are not defined at $0$ and are excluded by the domain of the statement.

The integer statement is proved with five routes and the sharp equality case: Route 1 is [L5] quoted in step 1.2; Route 2 is the geometric-sum comparison of step 1.3; Route 3 is the supporting-line remainder of step 1.4, whose identity has positive coefficients and double zero exactly at $y=1$; Route 4 is the AM-GM application of step 1.5; and Route 5 is the binomial expansion, which gives strictness for $x>0$ in step 2.1 and for $-1\le x<0$ in step 2.2; all routes cover $x\ge -1$, the degenerate equalities $n\le 1$ or $x=0$ are those of step 1.1, and every route gives strict inequality for $n\ge 2$, $x\mathrm{\ne }0$.

The rational forms hold on the stated domain: for $x>-1$ the inequalities are the three cases of step 1.8, their equality and strictness pattern is step 2.3, and the endpoint $x=-1$ is step 2.4; hence $(1+x{)}^r\ge 1+rx$ for $r\ge 1$ or $r\le 0$ and $(1+x{)}^r\le 1+rx$ for $0\le r\le 1$, in each case exactly on the stated domain, with equality exactly when $x=0$ or $r\in \{0,1\}$.

The quadratic refinement: let $r\ge 2$ be rational and $x\ge -1$. If $x>-1$, put $y:=1+x>0$, $z:=y^2>0$ and $s:=r\mathrm{/}2$, a rational with $s\ge 1$; the rational Bernoulli lower bound of step 3.2 applied to $z$ and $s$ gives $z^s\ge 1+s(z-1)$, and $z^s=(y^2{)}^{r\mathrm{/}2}=y^r$ while $s(z-1)=(r\mathrm{/}2)((1+x{)}^2-1)=(r\mathrm{/}2)(2x+x^2)$, so $y^r\ge 1+rx+(r\mathrm{/}2)x^2$; equality in this chain holds exactly when $z=1$, that is $x=0$, or $s=1$, that is $r=2$; if $x=-1$ the left side is $0^r=0$ and the right side is $1-r+r\mathrm{/}2=1-r\mathrm{/}2\le 0$ with equality exactly at $r=2$; and if $m$ is a natural with $2m\le r$, then $r\mathrm{/}2-m\ge 0$ and $x^2\ge 0$ give $(r\mathrm{/}2)x^2\ge m{x}^2$, so the stronger bound implies $(1+x{)}^r\ge 1+rx+m{x}^2$; for $1\le r<2$ there is no positive natural $m$ with $2m\le r$, the coefficient $r\mathrm{/}2$ lies below $1$, and the ordinary lower bound is the case $m=0$.

Summary: the integer Bernoulli statement holds with the equality clause $n\le 1$ or $x=0$ by five independent finite routes, two of which (Routes 4 and 5) are new here while Route 1 is the published induction and Routes 2 and 3 are the difference-of-powers and supporting-line arguments (steps 1.2 through 1.5, 2.1, 2.2, 3.1); the rational-exponent forms hold with the equality clause $x=0$ or $r\in \{0,1\}$ including the directly verified endpoint (steps 1.8, 2.3, 2.4, 3.2); and the quadratic refinement $(1+x{)}^r\ge 1+rx+(r\mathrm{/}2)x^2$ for $r\ge 2$, holds with equality exactly when $x=0$ or $r=2$ (step 4.1), which also proves the coefficient weakening and the ordinary bound for $1\le r<2$ without assigning them that equality clause; no route invokes a derivative, logarithm, exponential, convergence, or an unproved optimization theorem.