(2/π)x ≤ sin(x) ≤ x for all x in [0,π/2]
Proof: sin's slope is positive (but never greater than 1) in the interval,
and concave down, so the graphs of
y=x, y=sin(x), y=(2/π)x touch only at the endpoints of the interval.
Given f, a strictly increasing function whose value is 0 at 0, the
integral from 0 to a of f plus the integral from 0 to b of f-1 is
greater than the product ab. This is fairly obvious from the area
interpretation of the integral, but not so obvious is that choosing f(x)=xp-1
results in this famous special case of Young's Inequality:
Let a, b > 0; p, q > 0; 1/p + 1/q = 1. Then ab ≤ ap/p +
Proof: log is concave down, so log ab = (1/p)log ap + (1/q)log bq
≤ log (1/p)ap + log (1/q)bq
Let x > -1, r > 1. Then (1+x)r ≥ 1+xr
For a,b,c>0, a/(b+c)+b/(a+c)+c/(a+b) ≥ 3/2
For x1,x2,...,xn > 0, n ≤ 12 or n is odd
and n ≤ 23, then
the sum xi/(xi+1+xi+2) ≥ n/2, where the
subscripts are understood modulo n.
For all non-negative real numbers x, y, z and a positive number t,
xt(x-y)(x-z) + yt(y-z)(y-x) + zt(z-x)(z-y)
If a1 ≥ a2 ≥ ... ≥ an and b1
≥ b2 ≥ ... ≥ bn
The proof is that the RHS can be written as the sum of n different
(a1b1 + a2b2 + ...
+ an-1bn-1 + anbn) +
(a1b2 + a2b3 + ...
+ an-1bn + anb1) +
(a1bn + a2b1 + ...
+ an-1bn-2 + anbn-1)
The first of these n sums is at least as big as each of the others by the
rearrangement inequality, and n times the LHS, so the LHS is at least as
big as the RHS.
MacLaurin's inequality is a generalization of the AM-GM inequality.
First, define the product of a set as the product of all the elements of
Next, define the MacLaurin Sum Sk as the average of all the
products of k-element subsets of an n-element set.
The MacLaurin inequality is S1 ≥ S21/2
≥ S31/3 ≥ ... ≥ Sn1/n
Thus the AM-GM inequality represents the endpoints of a longer chain of
Let a1, a2, ... be a sequence of real numbers.
Carleman's inequality states that
|(a1a2...an)1/n ≤ e
Arithmetic-geometric-harmonic means inequality
Let x1, x2, ..., xn be positive numbers.
Max(x1, x2, ..., xn) ≥
(x1 + x2 + ... + xn)/n (the
arithmetic mean) ≥
(x1 x2 ... xn)1/n (the
geometric mean) ≥
n/(1/x1 + 1/x2 + ... + 1/xn) (the harmonic
Min(x1, x2, ..., xn),
(with equality only if x1=x2=...=xn)
General Means inequality, a.k.a. Power Means inequality
The power means inequality is a generalization of the AM-GM-HM
Define Mr, the rth power mean of nonnegative
numbers a1, a2, ... an as
When r ≠ 0, Mr(a1, a2, ... an)
= Average(a1r, a2r, ... anr)1/r
The limit of Mr, as r approaches zero, is the Geometric Mean,
when r=0, Mr(a1, a2, ... an) =
a2 ... an)1/n, the Geometric Mean.
The Power Means inequality states that whenever x<y, Mx(a1,
a2, ... an) ≤ My(a1, a2,
(with equality only if a1=a2=...=an)
If f is a convex function, then the value of the function at the weighted
average of selected values of its domain is less than or equal to the
weighted average (same weights) of the values of the function of those same
selected values of its domain.
If f is a concave function (sometimes called convex down) then the
inequality is reversed.
If x1, x2, ..., xn and y1, y2,
..., yn are two sequences of positive numbers, then the sum
x1y1 + x2y2 + ... + xnyn
is maximized when the two sequences are ordered the same way (i.e. x1
≤ x2 ≤ ... ≤ xn and y1 ≤ y2 ≤
... ≤ yn) and is minimized when the two sequences are ordered
opposite to one another.
Let a1, ..., an; b1, ..., bn;
... ; z1, ..., zn be sequences of nonnegative real
numbers, and let la,
lb, ..., lz be
positive reals which sum to 1. Then:
(a1 + ... + an)la(b1
+ ... + bn)lb...(z1
+ ... + zn)lz
+ ... + anlabnlb...znlz
Cauchy's inequality is a special case of this in which n=2 and
= lb = 1/2.