Examples of inequality implied by equality.

Question

It is well known Cauchy's inequality is implied by Lagrange's identity. Bohr's inequality $|a -b|^2 \le p|a|^2 +q|b|^2$, where $\frac{1}{p}+\frac{1}{q}=1$, is implied by $|a -b|^2 +|\sqrt{p/q}a+\sqrt{q/p}b|^2= p|a|^2 +q|b|^2$. L.K Hua's determinant inequality $\det(I-A^*A)\cdot \det(I-B^*B)\le |\det(I-A^*B)|^2$ is implied by Hua's matrix equality $(I-B^*B)-(I-B^*A)(I-A^*A)^{-1}(I-A^*B)=-(A-B)^*(I-AA^*)(A-B)$. What other examples can be found?

Answer 1

Lots of number theoretic inequalities are to be had from the binomial theorem. I remember reading the below argument as part of Erdos's proof of Bertrand's postulate:

Suppose that $n$ is a positive integer, then we have

$$4^n = (1 + 1)^{2n} = {\sum_{j=0}^{2n}}{2n\choose{j}}.$$

Thus, since $ 2n\choose{n}$ is the maximum value of the sequence $({2n\choose{k}})$, we conclude that

$$ 4^n < (2n + 1){2n\choose{n}}.$$

I thought it was neat.

Answer 2

Here are two very elementary examples.

It's not 100% different from your Cauchy's inequality example, but the fact that if X is a random variable, then $(\mathbb{E}X)^2\leq\mathbb{E}X^2$ is very useful and follows from the fact that the difference equals the variance of X.

The fact that $|\cos x|\leq 1$ and $|\sin x|\leq 1$ follows from the fact that $\cos^2x+\sin^2x=1$.

Answer 3

In an arbitrary triangle whose circumcircle has radius $R$ and center $O$ and whose inscribed circle has radius $r$ and center $I$, we have Euler's inequality $$R\geq 2r$$ This follows from the equality $$|IO|^2=R(R-2r)$$ (There are many examples in Euclidean geometry, I think Ptolemy's inequality follows from an equality but I can't remember at the moment)

Answer 4

After A1 from the 1968 Putnam:

$$\frac {22}7 - \pi = \int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx \gt 0$$

Integral proofs that $355/113 \gt \pi$.

I expect that there should be a proof of Jensen's inequality as an integral of a nonnegative quantity.

Answer 5

Over real-closed fields such as $\langle \mathbb{R}, +, *, -, <, 0, 1 \rangle$, there is an interesting simple answer: every polynomial inequality is equivalent to a projected equation. E.g., Given $p_1, p_2 \in \mathbb{Q}[\vec{x}]$ we have $\left( p_1 > p_2 \ \iff \ \exists z \text{ s.t. } z^2(p_1 - p_2) - 1 = 0 \right),$ and $\left( p_1 \geq p_2 \ \iff \ \exists z \text{ s.t. } p_1 - p_2 - z^2 = 0 \right).$

Geometrically, this is the simple observation that every semialgebraic set defined as the set of $n-$dimensional real vectors satisfying an inequality is the projection of an $n+1$-dimensional real-algebraic variety defined by a single equation. Semialgebraic sets defined by boolean combinations of equations and inequalities can be similarly encoded as the set of satisfying real vectors of (an) equation(s) by using the Rabinowitsch encoding $(p_1 = 0 \vee p_2 = 0 \ \iff p_1p_2 = 0)$ and $(p_1 = 0 \wedge p_2 = 0 \ \iff p_1^2 + p_2^2 = 0).$

Combining the above two observations, one obtains the fact that every semi-algebraic set $S \subseteq \mathbb{R}^n$ is the projection of a real algebraic variety $V \subseteq \mathbb{R}^{n+k}$, where $k$ is the number of inequality symbols appearing in the defining Tarski formula for $S$. In fact, due to a construction of Motzkin [``The Real Solution Set of a System of Algebraic Inequalities is the Projection of a Hypersurface in One More Dimension,'' Inequalities II, O. Shisha, ed., 251-254, Academic Press (1970)], it is known that every such $S$ is in fact the projection of a real-algebraic variety in $\mathbb{R}^{n+1}$.

Answer 6

Two examples due to Hurwitz.

• The AM-GM inequality. For the function $f=f(x_1,x_2,\dots,x_n)$ let $Pf(x_1,x_2,\dots,x_n)$ denote the sum of $f$ over the $n!$ quantities that result from all possible $n!$ permutations of the $x_i$. Then $$\frac{x_1^n+x_2^n+\dots+x_n^n}{n}-x_1x_2\dots x_n=\frac{1}{2\ n!}(\phi_1+\phi_2+ \dots \phi_n),$$ where $$\phi_k=P[(x_1^{n-k}-x_2^{n-k})(x_1-x_2)x_3x_4\dots x_{k+1}]=P[(x_1-x_2)^2(x_1^{n-k-1}+\dots x_2^{n-k-1})x_3x_4\dots x_{k+1}]\geq0.$$ The proof can be found in Inequalities by Beckenbach and Bellman.

• The isoperimetric inequality. Let the boundary of $\Omega\subset \mathbb R^2$ be a rectifiable Jordan curve $\partial \Omega=\{((x(s),y(s))|\ s\in[0,2\pi))\}$. Then $$L^2-4\pi A=2\pi^2\sum\limits_{n=1}^{\infty}\left[(na_n-d_n)^2+(nb_n+c_n)^2+ (n^2-1)(c_n^2+d_n^2)\right],$$ where $$x(s)=\sum\limits_{n=0}^{\infty}(a_n\cos ns+b_n\sin ns),\quad y(s)=\sum\limits_{n=0}^{\infty}(c_n\cos ns+d_n\sin ns).$$

Answer 7

Another "Hilbertian" example: Bessel's inequality follows from Bessel's equality. See, e.g., http://www.math.uri.edu/~quinn/web/mth629_Bessels.pdf.

And now (maybe off-topic, but the question is rather vague) an example of an inequality derived via an identity:

The (simple) identity is the so-called "multiplication of means", roughly: the expectation of a product of independent random variables equals the product of their expectations. The (not so simple) inequality is the Grothendieck one: http://www.ams.org/proc/1987-100-01/S0002-9939-1987-0883401-0/S0002-9939-1987-0883401-0.pdf. (Well, it is not obtained from that identity in an obvious and direct way, but the identity is an essential ingredient in the proof.)

Answer 8

Here are some more elementary examples.

• The easier cases of the AM-GM inequality follow from equalities, namely $a^2+b^2\geq 2ab$ because $a^2+b^2-2ab=(a-b)^2$ and $\frac{a^3+b^3+c^3}{3}\geq abc$ for $a,b,c\geq 0$ because $a^3+b^3+c^3-3abc=\frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$.

• We have that for any triangle ABC the point X that minimizes $AX^2+BX^2+CX^2$ is the centroid G because of Leibniz's relation $AX^2+BX^2+CX^2=AG^2+BG^2+CG^2+3XG^2$.