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

7$\begingroup$ This should be community wiki. $\endgroup$ – Steve Huntsman Mar 8 '10 at 16:13

3$\begingroup$ For polynomial inequalities there is Hilbert's 17th problem: en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem $\endgroup$ – Qiaochu Yuan Mar 8 '10 at 16:18

1$\begingroup$ In view of Harald's comment, maybe it is also interesting to ask for examples of inequalities which are not implied by equalities. $\endgroup$ – Pete L. Clark Mar 8 '10 at 19:35

1$\begingroup$ IMHO, one of the greatest inequalities is the Jensen´s one. Lots of inequalities follow from it. But is it implied by any equality? $\endgroup$ – M.G. Mar 8 '10 at 19:59

1$\begingroup$ It's hard to formalize Pete's question, since if A is always at most B then BA=C for some C that is always nonnegative. One needs a notion of "obviously nonnegative", which is often provided by a number's being a sum of squares. $\endgroup$ – gowers Mar 8 '10 at 20:03
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.
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 $\cos^2x+\sin^2x=1$.

$\begingroup$ Your second example is restricted in real field. My examples are also true in complex field. $\endgroup$ – Sunni Mar 8 '10 at 21:26

$\begingroup$ Isn't your first example Cauchy's inequality applied to the scalar product on $L^2$ of the probability space? $\endgroup$ – Andrea Ferretti Mar 8 '10 at 21:51

$\begingroup$ It's CauchySchwarz applied to the scalar product of X and the constant function 1. $\endgroup$ – gowers Mar 8 '10 at 23:55

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 $$IO^2=R(R2r)$$ (There are many examples in Euclidean geometry, I think Ptolemy's inequality follows from an equality but I can't remember at the moment)
After A1 from the 1968 Putnam:
$$\frac {22}7  \pi = \int_0^1 \frac{x^4(1x)^4}{1+x^2}dx \gt 0$$
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.

$\begingroup$ Your equality cannot imply $355/113>\pi$, because 22/7 = 3.1429, 355/113 = 3.1416 $\endgroup$ – Sunni Mar 9 '10 at 14:04

$\begingroup$ +1: Cool! I'm no Putnam connoisseur (on the contrary), but the 1968 specimen looks to be much more interesting than usual. Can anyone confirm/deny this? $\endgroup$ – Pete L. Clark Mar 9 '10 at 14:07

1$\begingroup$ @miwalin The idea is to construct an analogous integral of something which is obviously positive which has the value $355/113\pi.$ A few examples were given in the paper linked. Given such an integral, we can compute better estimates for $\pi$ using rigorous numerical integration techniques. For example, we can use Simpson's Rule whose error is bounded by the magnitude of the 4th derivative of the integrand on the interval, and we can bound that. @Pete Yes, the problems on that competition seem particularly nice, although that's subjective. $\endgroup$ – Douglas Zare Mar 9 '10 at 15:06

1$\begingroup$ Since the link in the answer no longer works, here is Internet Archive snapshot from 2010. It seems that some copies of the paper can be found online  Google, Google Scholar  perhaps at least some of these links will last. $\endgroup$ – Martin Sleziak Sep 3 '18 at 15:37
Over realclosed 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 realalgebraic 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 semialgebraic 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., 251254, Academic Press (1970)], it is known that every such $S$ is in fact the projection of a realalgebraic variety in $\mathbb{R}^{n+1}$.
Two examples due to Hurwitz.
The AMGM 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 $$\phi_k=P[(x_1^{nk}x_2^{nk})(x_1x_2)x_3x_4\dots x_{k+1}]=P[(x_1x_2)^2(x_1^{nk1}+\dots x_2^{nk1})x_3x_4\dots x_{k+1}]\geq0.$$ 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 $$L^24\pi A=2\pi^2\sum\limits_{n=1}^{\infty}\left[(na_nd_n)^2+(nb_n+c_n)^2+ (n^21)(c_n^2+d_n^2)\right],$$ where $$x(s)=\sum\limits_{n=0}^{\infty}(a_n\cos ns+b_n\sin ns),\quad y(s)=\sum\limits_{n=0}^{\infty}(c_n\cos ns+d_n\sin ns).$$

$\begingroup$ Very interesting. This is not included in Hardy, Littlewood and Polya's 'Inequalities'. $\endgroup$ – Sunni Apr 27 '10 at 3:49
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 offtopic, but the question is rather vague) an example of an inequality derived via an identity:
The (simple) identity is the socalled "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/198710001/S00029939198708834010/S00029939198708834010.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.)

$\begingroup$ Bessel's inequality is also implied by Parseval's identity (given an orthonormal set, extend it to an orthonormal basis, apply Parseval, and throw away the unwanted terms). $\endgroup$ – Nate Eldredge Apr 26 '10 at 15:34
Here are some more elementary examples.
The easier cases of the AMGM inequality follow from equalities, namely $a^2+b^2\geq 2ab$ because $a^2+b^22ab=(ab)^2$ and $\frac{a^3+b^3+c^3}{3}\geq abc$ for $a,b,c\geq 0$ because $a^3+b^3+c^33abc=\frac{1}{2}(a+b+c)((ab)^2+(bc)^2+(ca)^2)$.
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 $AX^2+BX^2+CX^2=AG^2+BG^2+CG^2+3XG^2$.

$\begingroup$ The second one is nice, while the first one is too elementary. $\endgroup$ – Sunni Mar 8 '10 at 23:47