_inequalities_for_inmotc_-2020

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Power mean inequality

Definition 1 Given a sequence x1 , x2 , . . . , xn of positive real numbers, the mean of order r , denoted by Mr (x) is defined as  Mr (x) =

xr1 + xr2 + · · · + xrn n

 r1

M1 (x1 , . . . , xn ) is the arithmetic mean, while M2 (x1 , . . . , xn ) is the quadratic mean of the numbers x1 , . . . , xn M−1 is the harmonic mean. M0 can’t be defined using the expression analogous to the expressions for other means, but we will show that as r approaches 0 , Mr will approach the geometric mean. The famous mean inequality can be now stated as Mr (x1 , . . . , xn ) ≤ Ms (x1 , . . . , xn ), for 0 ≤ r ≤ s. However we will treat this in slightly greater generality. We will consider the weighted mean of order r . Definition 2 (Weighted mean) Let m = (m1 , . . . , mn ) be a fixed sequence of non-negative real numbers such that m1 + m2 + · · · + mn = 1 Then the weighted mean of order r of the sequence of positive reals x = (x1 , . . . , xn ) is defined as: 1

Mrm (x) = (m1 xr1 + m2 xr2 + · · · + mn xrn ) r

P ∴ M1m = ( mi xi ) =AM ∴ If m1 = m2 = · · · = n1 then M m r (x) = Mr (x) where Mr (x) is previously defined.

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Theorem 1 (General Mean Inequality) If x = (x1 , . . . , xn ) is a sequence of positive real numbers and m = (m1 , . . . , mn ) another sequence of positive real numbers satisfying m1 + m2 + · · · + mn = 1 , then for 0 ≤ r ≤ s we haveMrm (x) ≤ Msm (x) . • If m1 = m2 = · · · = Problems

2

1 n

1

then Mrm (x) = Mr (x) = (m1 xr1 + m2 xr2 + · · · + mn xrn ) r

Inequality of Minkowski

The standard inequalities are easily obtained by placing mi = 1 whenever some m appears in the text below. Assuming that the sum m1 +· · ·+mn = 1 one easily get the generalized (weighted) mean inequalities, and additional assumption mi = 1/n gives the standard mean inequalities. Theorem 2(Minkowski) If x1 , x2 , · · ·, xn ; y1 , y2 , · · ·, yn , and m1 , m2 , · · ·, mn are three sequences of  n  p1  n  p1 P P p p positive real numbers and p > 1 , then (xi + yi ) mi ≤ xi m i + i=1 i=1  n  p1 P p − − − − − (3) yi m i i=1

The equality holds if and only if the sequences (xi ) and (yi ) are proportional, i.e. if and only if there is a constant λ such that xi = λyi for 1 ≤ i ≤ n.

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Young’s inequality

If a, b, p , and q are positive real numbers such that Equality holds if and only if ap = bq

1 1 p+q

= 1 , then

ap p

q

+ bq ≥ ab

Lemma If x1 , x2 , . . . , xn , y1 , y2 , . . . , yn , m1 , m2 , . . . , mn are three sequences of positive real numbers and p, q > 1 such that p1 + 1q = 1, and α > 0 , then n X i=1

xi yi mi ≤

n n 1 pX p 1 X q α xi m i + q y mi − − − − − −(4) p qα i=1 i i=1

The equality holds if and only if αp xpi =

2

yiq αq

for 1 ≤ i ≤ n.

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Holders inequality • If x1 , x2 , . . . , xn , y1 , y2 , . . . , yn , m1 , m2 , . . . , mn are three sequences of positive real numbers and p, q > 1 such that p1 + 1q = 1, and α > 0 , then n X

xi yi mi ≤

i=1

n X

! p1

n X

xpi mi

! q1 yiq mi

− − − − − −(5)

i=1

i=1

The equality holds if and only if the sequences (xpi ) and (yiq ) are proportional.  n 2  n  n  P P 2 P 2 • For p = q = 2 we get CS xi yi ≤ xi yi . i=1

i=1

i=1

• The easy version for n = 3 , is for positive real numbers a, b, c, p, q, r, x, y, z, (x3 + y 3 + z 3 )(a3 + b3 + c3 )(p3 + q 3 + r3 ) ≥ (xap + ybq + acr)3

Problems 1. For positive integer n, define Sn to be the minimum value of the sum n q X (2k − 1)2 + a2k , k=1

where a1 , a2 , . . . , an are positive real numbers whose sum is 17. There is a unique positive integer n for which Sn is also an integer. Find this n. 2. Prove that for all positive real numbers a, b, c, x, y, z,

a3 x

+

b3 y

+

c3 z

≥

(a+b+c)3 3(x+y+z)

3. √ For nonnegative real numbers x, y, z prove that 3z 2 + zx ≤ 2(x + y + z).

p p 3x2 + xy + 3y 2 + yz +

4. Let a, b, c be positive real numbers such that a + b + c = 1.Prove the inequality a b c √ +√ +√ ≥1 c + 2a a + 2b b + 2c

5. (IMO 2001) Prove that for all positive real numbers a, b, c, √

a b c +√ +√ ≥ 1. a2 + 8bc b2 + 8ca c2 + 8ab

6. For a, b, c > 0, λ ≥ 8, prove that √

a2

a b c 3 +√ +√ ≥√ . 2 2 1+λ + λbc b + λac c + λab 3

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Cauchy flipping 1. Let a, b, c be non negatives with a2 +b2 +c2 = 3. Show that

P

cyc

1.

a3

1 ≥ +2

2. Let a, b, c be positive real numbers with the sum 3 . Prove that a b c 3 + + ≥ (Bulgaria TST 2003) 1 + b2 1 + c2 1 + a2 2 3. Suppose that a, b, c, d are four positive real numbers with sum 4. Prove that X a ≥2 1 + b2 c 4. (Pham Kim Hung) Let a, b, c be positive real numbers. Prove that 3 3 3 3 a) a2a+b2 + b2b+c2 + c2c+d2 + d2d+a2 ≥ a+b+c+d 2 4

a b) a3 +2b 3 +

b4 b3 +2c3

+

c4 c3 +2d3

+

d4 d3 +2a3

≥

a+b+c+d 3

5. Let a, b, c be positive real numbers with sum 3. Prove that a2 b2 c2 + + ≥1 2 2 a + 2b b + 2c c + 2a2 6. Let a, b, c be positive real numbers with sum 3. Prove that a2 b2 c2 + + ≥1 a + 2b3 b + 2c3 c2 + 2a3

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Schur’s Inequality January 17, 2020 Let a, b, c be non negative real numbers, and p be positive. Then the following inequality holds: ap (a − b)(a − c) + bp (b − c)(b − a) + cp (c − a)(c − b) ≥ 0, with equality if and only if a = b = c or a = b, c = 0 (up to permutation). The above inequality is known as Schur’s inequality, after Issai Schur. Common Cases • The n = 1 case yields the well-known inequality: a3 + b3 + c3 + 3abc ≥ a2 b + a2 c + b2 a + b2 c + c2 a + c2 b • The n = 2, an equivalent form is: a4 + b4 + c4 + abc(a + b + c) ≥ a3 b + a3 c + b3 a + b3 c + c3 a + c3 b • Corollary Let x, y, z and a, b, c be positive real numbers such that a ≥ b ≥ c or a ≤ b ≤ c. Then we have a(x − y)(x − z) + b(y − x)(y − z) + c(z − x)(z − y) ≥ 0.

Generalization of Schur’s lemma: Let a, b, c be positive real numbers,p, q ∈ R and q be positive. Then the following inequality holds: ap (aq − bq )(aq − cq ) + bp (bq − cq )(bq − aq ) + cp (cq − aq )(cq − bq ) ≥ 0 with equality if and only if a = b = c or a = b, c = 0 (up to permutation). 5

Generalized Form It has been shown by V alentinV ornicu that a more general form of Schur’s Inequality exists. Consider a, b, c, x, y, z ∈ R, where a ≥ b ≥ c, and either x ≥ y ≥ z or z ≥ y ≥ x. Let k ∈ Z+ , and let f : R → R+ 0 be either convex or monotonic. Then, f (x)(a − b)k (a − c)k + f (y)(b − a)k (b − c)k + f (z)(c − a)k (c − b)k ≥ 0. with equality if and only if a = b = c or a = b, c = 0 (up to permutation). The standard form of Schur’s is the case of this inequality where x = a, y = b, z = c, k = 1, f (m) = mr .

Problems 1. Let a, b, c be positive real numbers. Prove the inequality X

X a2 + bc a ≥ b+c (a + b)(a + c)

2. Let a, b, c be non-negative real numbers. Prove the inequality 4b2

b c 1 a + 2 + 2 ≥ 2 2 2 + bc + 4c 4c + ca + 4a 4a + ab + 4b a+b+c

3. (IMO 1964) Let a, b, c be the side lengths of a triangle. Prove that a2 (b + c − a) + b2 (c + a − b) + c2 (a + b − c) ≤ 3abc. Muirhead’s Inequality • Muirhead’s Inequality states that if a sequence a majorizes a sequence b, then a set of positivePreals x1 , x2 , · · · , xn : P given a1 a2 x x · · · xn an ≥ sym x1 b1 x2 b2 · · · xn bn 1 2 sym that is a  b implies T [a] ≥ T [b] . Equality is achieved only if either a = b, or all x0s are equal. 1. (BMO1 2002 Round 1): For positive real x, y, z such that x2 + y 2 + z 2 = 1prove that 1 x2 yz + xy 2 z + xyz 2 ≤ 3 2. (BMO1 1996 Round 1) Let a, b and c be positive real numbers. Prove that 4(a3 + b3 ) ≥ (a + b)3 and

9(a3 + b3 + c3 ) ≥ (a + b + c)3

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3. Let a, b, c be positive real numbers. Prove the inequality 1 1 1 b3 +c3 +abc + a3 +c3 +abc ≤ abc

1 a3 +b3 +abc

4. Let x, y and z be positive real numbers such that xyz ≥ 1. Prove that y5 − y2 z5 − z2 x5 − x2 + + ≥0 x5 + y 2 + z 2 y 5 + z 2 + x2 z 5 + x2 + y 2

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