Roots, Rational Powers, and Classical Inequalities: Examples and Counterexamples

At the witness, $1+x=1+(-4)=-3$ and $(-3{)}^3=(-3{)}^2\cdot (-3)=9\cdot (-3)=-(9\cdot 3)=-27$ by the recursion clause of [L1] and the sign rules of [L2]; so $(1+x{)}^3=-27$.

Also $3x=3\cdot (-4)=-(3\cdot 4)=-12$ by [L2], and $1+3x=1+(-12)=-11$ by field arithmetic; so the right-hand side of the claimed inequality is $-11$ at the witness.

Boundary evaluations: at $n=0$ both sides are $1$; at $n=1$ the recursion clause gives $(1+x{)}^1=(1+x{)}^0(1+x)=1+x=1+1\cdot x$ for every $x$; at $x=-3$, $n=3$ one gets $(1+x{)}^3=(-2{)}^3=-8$ and $1+3x=1+(-9)=-8$, so the two sides agree there too, at a point with $x<-1$; and at $x=-1$ with $n\ge 2$ the left side is $0$ while $1+nx=1-n<0$.

The comparison, using the evaluations of step 1.1 and step 1.2: $0<16$, because $1>0$ gives $0<1$, repeatedly adding constants by [L3] gives $0<4$, and $4\cdot 4=16>0$ is a product of positives by [L2]; adding $c=-27$ to $0<16$ by [L3] gives $-27<16+(-27)=-11$.

Hence $(1+x{)}^3=-27<-11=1+3x$ at $x=-4$, $n=3$, so the conclusion of the inequality fails there; since $-4<-1$, the hypothesis $x\ge -1$ of [L4] cannot be dropped, and the evaluations of step 1.3 show that equality holds at $n=0$, at $n=1$, and at the example $x=-3$, $n=3$; these evaluations do not classify all cases with $x<-1$.

The two-entry evaluations of [L1] give $A=a_0^2+a_1^2$, $B=b_0^2+b_1^2$, $C=a_0{b}_0+a_1{b}_1$ and $D=a_0{b}_1-a_1{b}_0$, all four being real numbers determined by the given data.

The identity: expanding both products by distributivity and collecting terms by [L2] and [L3], $AB-C^2=a_0^2{b}_0^2+a_0^2{b}_1^2+a_1^2{b}_0^2+a_1^2{b}_1^2-(a_0^2{b}_0^2+2(a_0{b}_1)(a_1{b}_0)+a_1^2{b}_1^2)$, and the outer terms cancel, leaving $a_0^2{b}_1^2-2(a_0{b}_1)(a_1{b}_0)+a_1^2{b}_0^2$, which is $(a_0{b}_1{)}^2-2(a_0{b}_1)(a_1{b}_0)+(a_1{b}_0{)}^2=(a_0{b}_1-a_1{b}_0{)}^2=D^2$; the sign of the cross term is the rule of [L4].

The square $D^2$ is nonnegative and vanishes only at $D=0$: if $D\mathrm{\ne }0$ then $D^2>0$ by [L5], and if $D=0$ then $D^2=0\cdot 0=0$ by [L5] and [L2].

Substituting step 1.2 into step 1.3 gives $AB-C^2=D^2\ge 0$, that is $C^2\le AB$; and equality $C^2=AB$ holds exactly when $D^2=0$, which by step 1.3 means exactly $D=0$, that is $a_0{b}_1=a_1{b}_0$.

The instance: at $a_0=1$, $a_1=2$, $b_0=3$, $b_1=4$ direct evaluation gives $A=1+4=5$, $B=9+16=25$, $C=3+8=11$ and $D=4-6=-2$, so that $5\cdot 25-121=4=(-2{)}^2$, in agreement with the identity of step 1.2.

The equality criteria agree: by [L6] applied at $n=2$, equality $C^2=AB$ holds if and only if the lists $(a_0,a_1)$ and $(b_0,b_1)$ are proportional in the symmetric sense; by step 2.1 the same equality holds if and only if $D=0$; hence $a_0{b}_1=a_1{b}_0$ holds exactly when the two-entry lists are proportional, which is the identification claimed between the local criterion and the clause of [L6].

Collecting the results: the identity of step 1.2 holds for all reals, step 2.1 gives the inequality with its equality criterion, step 3.1 identifies that criterion with the two-entry proportionality clause of [L6], and step 2.2 verifies the numerical instance, completing the example.

At the witness: $ab=(-1)(-4)=4$ by [L5], so $G=4^{1\mathrm{/}2}=2$ by [L2]; the arithmetic mean is $A=(-1+(-4))\mathrm{/}2=-5\mathrm{/}2$; and $2>-5\mathrm{/}2$, because $-5\mathrm{/}2<0$ by [L3] (using $2(-5\mathrm{/}2)=-5<0$ and the equivalence $u<0⟺2u<0$, where $2,5>0$ by [L6] and [L4]) while $2>0$ by [L6], so transitivity in [L4] gives $G=2>-5\mathrm{/}2=A$; thus the claimed inequality fails at $(-1,-4)$.

For arbitrary negative entries $a,b<0$: the product is positive, $ab>0$, by [L3]; the root $G=(ab{)}^{1\mathrm{/}2}$ is then defined and positive by [L2]; and $a+b<0$ by adding the inequalities $a<0$ and $b<0$ in [L4], so $2A=a+b<0$, while $A\ge 0$ would give $2A\ge 0$ because a product of nonnegatives is nonnegative by [L3] and [L5]; hence $A<0$, and transitivity gives $G>A$; so the root form fails for every pair of negative entries, not only at the displayed witness.

The second failure mode: at $a=-1$, $b=1$ the product is $ab=-1$, so $G=(-1{)}^{1\mathrm{/}2}$ would be a real square root of $-1$, which does not exist by [L7]; the conclusion of the claimed inequality cannot even be stated there.

The squared form survives for all real $a,b$: with $A=(a+b)\mathrm{/}2$ one has $A^2-ab=((a+b{)}^2-4ab)\mathrm{/}4=(a-b{)}^2\mathrm{/}4=((a-b)\mathrm{/}2{)}^2\ge 0$ by [L5]; hence $ab\le A^2$ for arbitrary reals, the two-entry squared form of [L1] without any sign hypothesis.

Collecting: the root form of the two-term arithmetic-geometric mean inequality fails outright at the witness (step 1.1, where $G=2>-5\mathrm{/}2=A$), fails systematically for every pair of negative entries (step 1.2), and can fail by undefinedness (step 1.3); what the hypothesis $a,b\ge 0$ buys is precisely the passage from the squared form $ab\le A^2$, which survives without it (step 1.4), to the root form $G\le A$, which needs $G$ to exist and $A$ to be nonnegative; the hypothesis is therefore load-bearing and cannot be dropped.

For all reals $x,y$, $((x+y)\mathrm{/}2{)}^2-xy=((x-y)\mathrm{/}2{)}^2\ge 0$, with equality exactly when $x=y$: expanding the square and collecting terms gives the identity, [L6] gives $(x-y{)}^2\ge 0$ with equality exactly at $x=y$, and translating by the constant $xy$ in [L7] yields $xy\le ((x+y)\mathrm{/}2{)}^2$, with equality exactly when $x=y$.

Doubling naturals: by [L9] there is a function $j\mapsto N_j$ on $\mathbb{N}$ with $N_0=1$ and $N_{j+1}=N_j+N_j$, and $N_j\ge 1$ and $N_j>j$ for every $j$: at $j=0$ this is $1>0$, and if $N_j>j$ then $N_j\ge j+1$ by [L10], so $N_{j+1}=N_j+N_j\ge (j+1)+1>j+1$ by [L10] and [L7].

Degenerate cases disposed of once for all routes. If $n=1$ then ${\sum }_{k<1}{a}_k=a_0$, $A=a_0$ and $G=a_0^{1\mathrm{/}1}=a_0$ by [L1] and [L3], so equality holds and the constant-list condition is automatic. If $a_j=0$ for some $j<n$ then ${\prod }_{k<n}{a}_k=0$ and $G=0^{1\mathrm{/}n}=0$ by [L1] and [L3], while $A\ge 0$ because $A$ is the sum of nonnegative terms scaled by the positive $n^{-1}$ ([L1], [L11]); hence $G=0\le A$ and $\prod =0\le A^n$ with equality exactly when $A=0$ (for $A\ge 0$, $A^n=0$ forces $A=0$ by [L2]), which happens exactly when ${\sum }_{k<n}{a}_k=0$, equivalently $a_k=0$ for every $k$ by [L1], and since some entry is $0$ that is exactly the constant-list case; so every list with a zero entry or of length $1$ is settled, and all routes below may assume $n\ge 2$ and $a_k>0$ for every $k<n$.

Monotonicity of finite sums and products, with strictness. If $x_k\le y_k$ for all $k<m$ then ${\sum }_{k<m}{x}_k\le {\sum }_{k<m}{y}_k$ and, when all entries are nonnegative, ${\prod }_{k<m}{x}_k\le {\prod }_{k<m}{y}_k$; if moreover $x_j<y_j$ for some $j<m$, all $x_k\ge 0$ and all $y_k>0$, then both inequalities are strict. The nonstrict product comparison follows by induction, splitting off the last factor and applying [L8], starting from $1\le 1$; the sum comparison is [L1]. For strict sums, split off the index $j$ and add its strict comparison to the nonstrict comparisons of the remaining sums. For strict products, if an $x_k$ is zero the lower product is zero and the upper product is positive by [L1] and [L6]. Otherwise both products of the remaining entries are positive; call them $P\le Q$. Then $x_{j}P<y_{j}P\le y_{j}Q$, by positive scaling and [L8].

Route 1, the published Cauchy forward-backward induction: [L4] states verbatim that for every natural $n\ge 1$ and nonnegative entries ${\prod }_{k<n}{a}_k\le A^n$, equivalently $G\le A$, with equality exactly when all entries are equal; its proof is the doubling inductive argument followed by the arithmetic-mean descent, so this route is the published theorem itself and needs no further work.

Route 2, sum-preserving replacement, as an induction. Claim $S(m)$: every list of $m$ strictly positive reals with sum $m$ has product at most $1$, with equality only for the constant list of ones. $S(1)$ is immediate. Assume $S(m)$ and let $b_0,\dots ,b_m>0$ have sum $m+1$; if all entries equal $1$ the product is $1$, so suppose the list is not constant; then some entry is below $1$ and some entry is above $1$, because a list all of whose entries are at least $1$, with one strictly larger, has sum strictly greater than $m+1$ by [step 1.4], while a list with all entries at most $1$, with one strictly smaller, has sum strictly less than $m+1$; pick such $u=b_i<1<b_j=v$ and replace the pair by $1$ and $w:=u+v-1>0$, which preserves the sum and strictly increases the pair product since $w-uv=(1-u)(v-1)>0$ by [L6]; dropping the new entry $1$ leaves $m$ positive entries of sum $m$, whose product is $wR$ with $R>0$ the product of the remaining entries, so $S(m)$ gives $wR\le 1$ and the original product satisfies $uvR<wR\le 1$ by [step 1.4] and [L7]; hence every nonconstant list of sum $m+1$ has product strictly below $1$ and equality in $S(m+1)$ occurs only for the constant list, completing the induction.

Route 3, product-preserving replacement, as an induction. Claim $T(m)$: every list of $m$ strictly positive reals with product $1$ has sum at least $m$, with equality only for the constant list of ones. $T(1)$ is immediate. Assume $T(m)$ and let $c_0,\dots ,c_m>0$ have product $1$; if the list is constant then every entry is $1$ and the sum is $m+1$ with equality, so suppose it is not constant; then some entry is below $1$ and some entry is above $1$, since a list with all entries at least $1$ and one strictly larger has product strictly greater than $1$ by [step 1.4], and a list with all entries at most $1$ and one strictly smaller has product strictly less than $1$; pick $u=c_i<1<c_j=v$ and replace the pair by $1$ and $uv$, which preserves the product and strictly decreases the sum because $u+v-(1+uv)=(1-u)(v-1)>0$ by [L6]; dropping the $1$ leaves $m$ positive entries of product $uvR=1$ (with $R>0$ the product of the remaining entries), whose sum is at least $m$ by $T(m)$, so the replaced list has sum at least $m+1$ and the original list, whose sum is strictly larger, has sum strictly greater than $m+1$; hence equality in $T(m+1)$ occurs only for the constant list, completing the induction.

Route 4, induction from the integer Bernoulli inequality. Assume the claim at a length $m\ge 1$, and let $c_0,\dots ,c_m>0$ be a list of length $m+1$ with mean $A$; put $B:=\left({\sum }_{k<m}{c}_k\right)\mathrm{/}m>0$, $c:=c_m>0$, $t:=c{B}^{-1}>0$ and $h:=(t-1)\mathrm{/}(m+1)$, so that $h>-1$ because $t>0$ and $m+1\ge 2$; the induction hypothesis gives ${\prod }_{k<m+1}{c}_k=\left({\prod }_{k<m}{c}_k\right)c\le B^{m}c=B^{m+1}t$ by [L1] and [L2], while Bernoulli's inequality [L5] at $h$ and $m+1$ gives $(1+h{)}^{m+1}\ge 1+(m+1)h=t$; since $B(1+h)=B+B(t-1)\mathrm{/}(m+1)=B+(c-B)\mathrm{/}(m+1)=(mB+c)\mathrm{/}(m+1)=A$ by [L1], [L11] and field arithmetic, the bound continues as $B^{m+1}t\le B^{m+1}(1+h{)}^{m+1}=(B(1+h){)}^{m+1}=A^{m+1}$, so the claim holds at length $m+1$; if ${\prod }_{k<m+1}{c}_k=A^{m+1}$ then both inequalities in the chain are equalities, and since $m+1\ge 2$ the equality clause of [L5] gives $h=0$, that is $t=1$ and $c=B$, whereupon $\left({\prod }_{k<m}{c}_k\right)B=B^{m}B$ cancels the nonzero factor $B$ through [L11] to give ${\prod }_{k<m}{c}_k=B^m$, so the equality clause of the induction hypothesis gives $c_0=\cdots =c_{m-1}=B=c$ and the list is constant; conversely a constant list has equality, and with step 1.3 supplying the base length $1$ this completes the induction.

Route 5a, the doubling step with its equality clause. Let $m\ge 1$ and assume $U(m)$: every nonnegative list of length $m$ has product at most the $m$-th power of its mean, with equality only when the list is constant. Let $a_0,\dots ,a_{m+m-1}$ be a list of length $m+m$, which by step 1.3 may be taken strictly positive; split the sums and products at $m$ by [L1], set $X:=\left({\sum }_{k<m}{a}_k\right)\mathrm{/}m$ and $Y:=\left({\sum }_{k=m}^{m+m-1}{a}_k\right)\mathrm{/}m$, both positive, and let $A$ be the total mean, so that $(m+m)A=mX+mY$ and hence $A=(X+Y)\mathrm{/}2$ by [L1] and [L11]; applying $U(m)$ to the two halves gives ${\prod }_{k<m+m}{a}_k=\left({\prod }_{k<m}{a}_k\right)\left({\prod }_{k=m}^{m+m-1}{a}_k\right)\le X^m{Y}^m=(XY{)}^m\le (A^2{)}^m=A^{m+m}$ by [L1], [L2] and the square identity of step 1.1; if equality holds throughout, injectivity in [L2] gives $XY=A^2$, hence $X=Y$ by step 1.1; furthermore each half-product must equal its bound, since a strict bound for one half, multiplied by the positive other half-product, would make the total product strictly less than $X^m{Y}^m$ by step 1.4; thus $U(m)$'s equality clause makes each half constant with value equal to its mean, so all $m+m$ entries are equal; conversely a constant list gives equality; zero entries and length-one lists are already covered by step 1.3, so $U(m+m)$ holds.

Conclusion of Route 2. Let $a_0,\dots ,a_{n-1}>0$ with $n\ge 2$ and mean $A>0$; the normalized entries $b_k:=a_k{A}^{-1}$ are positive and have sum $\left({\sum }_{k<n}{a}_k\right)A^{-1}=nA{A}^{-1}=n$ by [L1] and [L11], so $S(n)$ of step 2.1 gives ${\prod }_{k<n}{b}_k\le 1$ with equality only when every $b_k=1$; since ${\prod }_{k<n}{b}_k=\left({\prod }_{k<n}{a}_k\right)A^{-n}$ by [L1] and [L2], multiplying by the positive $A^n$ recovers ${\prod }_{k<n}{a}_k\le A^n$, with equality exactly when every $a_k=A$, that is exactly when the list is constant; and since $G^n={\prod }_{k<n}{a}_k$ with $G,A\ge 0$, the comparison is equivalent to $G\le A$ with the same equality case by [L2] and [L3].

Conclusion of Route 3. Let $a_0,\dots ,a_{n-1}>0$ with $n\ge 2$ and geometric mean $G>0$; the entries $c_k:=a_k{G}^{-1}$ are positive with product $\left({\prod }_{k<n}{a}_k\right)G^{-n}=G^n{G}^{-n}=1$ by [L1] and [L2], so $T(n)$ of step 2.2 gives ${\sum }_{k<n}{c}_k\ge n$ with equality only when every $c_k=1$, that is $\left({\sum }_{k<n}{a}_k\right)G^{-1}\ge n$ by [L1]; multiplying by the positive $G$ gives ${\sum }_{k<n}{a}_k\ge nG$, hence $A\ge G$, with equality exactly when every $a_k=G$, the constant-list case, and a constant list plainly attains equality.

Conclusion of Route 4. Step 1.3 disposes of the base length $1$, and step 2.3 derives the claim at length $m+1$ from length $m$ for every $m\ge 1$; hence by [L9] the statement ${\prod }_{k<n}{a}_k\le A^n$ with equality exactly for constant lists holds at every natural $n\ge 1$ by this route alone, which uses the integer Bernoulli inequality [L5] and never the arithmetic-geometric mean theorem.

Conclusion of Route 5. Let $a_0,\dots ,a_{n-1}>0$; step 1.2 supplies a doubling length $N:=N_n$ with $N\ge n$, and we pad with the geometric mean by setting $b_k:=a_k$ for $k<n$ and $b_k:=G$ for $n\le k<N$, so that ${\prod }_{k<N}{b}_k=\left({\prod }_{k<n}{a}_k\right)G^{N-n}=G^n{G}^{N-n}=G^N$ and ${\sum }_{k<N}{b}_k={\sum }_{k<n}{a}_k+(N-n)G$ by [L1] and [L2]; steps 1.1, 1.2 and 2.4 give $U(N)$, so $G^N={\prod }_{k<N}{b}_k\le \left(\right({\sum }_{k<n}{a}_k+(N-n)G)\mathrm{/}N{)}^N$ and hence $G\le ({\sum }_{k<n}{a}_k+(N-n)G)\mathrm{/}N$ by [L2]; multiplying by $N$ and subtracting the constant $(N-n)G$ in [L7] gives $nG\le {\sum }_{k<n}{a}_k$, that is $G\le A$; equality $G=A$ holds exactly when the padded list is constant by $U(N)$'s equality clause and [L2], and the padded list is constant exactly when all original entries equal $G$, which is the constant-list case.

The comparison of the five routes. Route 1 is the published Cauchy forward-backward proof quoted as [L4] in step 1.5; Route 2 is the normalized sum-preserving induction of step 2.1 with its translation in step 3.1; Route 3 is the normalized product-preserving induction of step 2.2 with its translation in step 3.2; Route 4 is the Bernoulli induction of step 2.3 with its base-and-induction assembly in step 3.3; Route 5 is the square identity of step 1.1, the doubling induction of step 2.4 and the geometric-mean padding of step 3.4, with no backward descent; each route establishes ${\prod }_{k<n}{a}_k\le A^n$, equivalently $G\le A$, with equality exactly for constant lists, and none of them uses a derivative, logarithm, exponential function, convergence or an optimization theorem.