Inequalities Theorems Kaimyn and Simon August

1 1.1

Basic and Important Ineqaulities Squares are Never Negative

Seems basic? Yes it is. This is the basic section. But, this would have to be one of the most important inequalities, and is used to build pretty much all of the others. So, remember this: 2 2 x+ 1 x2 + · · · + xn ≥ 0

With equality if and only if x1 = x2 = · · · = xn = 0

1.2

QM-AM-GM-HM

Let a1 , a2 , · · · , an be n positive reals. Then: r √ a21 + a22 + · · · + a2n a1 + a2 + · · · + an ≥ ≥ n a1 a2 · · · an ≥ n n

1 a1

+

1 a2

n + ··· +

1 an

Equality occurs if and only if a1 = a2 = · · · = an

1.3

The Rearrangement Inequality

Let a1 ≥ a2 ≥ · · · ≥ an be some list of real numbers and β1 , β2 , · · · , βn be some permutation of real numbers b1 ≥ b2 ≥ · · · bn . Then: a1 bn + a2 bn−1 + · · · + an b1 ≤ a1 β1 + a2 β2 + · · · + an βn ≤ a1 b1 + a2 b2 + · · · + an bn Equality occurs if and only if βi = βi+1 for all i such that ai < ai+1 . If a1 > a2 > · · · > an then equalitiy occurs when {β1 , β2 , · · · , βn } = {b1 , b2 , · · · , bn } or {bn , bn−1 , · · · , b1 }

1.4

Cauchy Schwarz

Let a1 , a2 , · · · , an and b1 , b2 , · · · , bn be two sets of positive reals. Then we have that: (a21 + a22 + · · · + a2n )(b21 + b22 + · · · + b2n ) ≥ (a1 b1 + a2 b2 + · · · + an bn )2 Equalitiy occurs if and only if

a1 b1

=

a2 b2

= ··· = 1

an bn

1.5

The Triangle Inequality

Let a, b, c > 0 be the sides of a triangle. Then it is a necessary and sufficient that a + b > c, b + c > a and c + a > b. Note that if we have, say, a + b = c then it is a degenerate triange: a line. This inequality is equivalent to making the substitution a = x + y, b = y + z. c = z + x. You can see that this substitution is valid for any triangle by considering the incircle and the “Ice Cream Cone Theorem.” You can also see that this substitution satisfies the triangle inequalities. Most geometric inequalities involve this in some way.

1.6

The Discrete Inequality

This one might belong more in number theory, but it’s worth mentioning here, because you do have a chance of coming across it. If a, b are integers, and a > b then we have that a ≥ b + 1.

2 2.1

No so basic inequalities Bernoulli’s Inequality

Suppose x > −1, x 6= 0 and n > 1 or n < 0. Then: (1 + x)n ≥ 1 + xn Alternatively, if 0 < x < 1 then the inequality is reversed.

2.2

Generalised Power Means

This is an extension of the QM-AM-GM-HM mentioned before. It states that, if α > β and a1 , a2 , · · · , an is some set of positive reals: ! β1 α1 α β β β a + a + · · · + a a1 + a2 α + · · · + aα 2 n n 1 ≥ n n Equality occurs if and only if a1 = a2 = · · · = an . Note that, in the case of α = 0 the limiting case of the geometric inequality occurs.

2.3

Weighted AM-GM

For a set a1 , a2 , · · · , an of non-negative reals, and ω1 , ω2 , · · · , ωn be a set of non-negative reals such that ω1 + ω2 + · · · + ωn = 1. Then: ωn 1 ω2 ω1 a1 + ω2 a2 + · · · + ωn an ≥ aω 1 a2 · · · an

2.4

Schur’s Inequality

Let x, y, z, n be non-negative real numbers. Then: xn (x − y)(x − z) + y n (y − z)(y − x) + z n (z − x)(z − y) ≥ 0 2

2.5

Jensen’s Inequality

NOTE: This one is a bit more advanced than the other ones. Knowing some calculus will help, but is not necessary. Let f (x) be a convex function. That is, f 00 (x) ≥ 0, or if we draw a line then it stays “inside” the graph. If you’re still confused about it, Google it, or look it up on Wikipedia. Infact, that’s pretty much how Jensen’s is defined: x1 + x2 f (x1 ) + f (x2 ) ≥f 2 2 If you draw a graph of a convex function this will be obvious. For a concave function, which is an upside-down convex function (f 00 (x) ≤ 0 ), we have the reverse, namely: f (x1 ) + f (x2 ) x1 + x2 ≤f 2 2 This can be extended to, for reals x1 , x2 , · · · , xn and weights ω1 +ω2 +· · ·+ωn = 1 to: ω1 f (x1 ) + ω2 f (x2 ) + · · · + ωn f (xn ) ≥ f (ω1 a1 + ω2 a2 + · · · + ωn an )

Convex

ω1 f (x1 ) + ω2 f (x2 ) + · · · + ωn f (xn ) ≤ f (ω1 a1 + ω2 a2 + · · · + ωn an )

Concave

2.6

Muirhead’s Inequality

Consider two sets of nonnegative reals: {α1 , α2 , · · · , αn } and {β1 , β2 , · · · , βn } such that α1 + α2 + · · · + αn = β1 + β2 + · · · + βn and α1 ≥ β1 α1 + α2 ≥ β1 + β2 .. . α1 + α2 + · · · + αn−1 ≥ β1 + β2 + · · · + βn−1 Then we have that: X

αn 1 α2 xα 1 x2 · · · xn ≥

sym

X

xβ1 1 xβ2 2 · · · xβnn

sym

Where x1 , x2 , · · · , xn are some non-negative real numbers.

3