Algebra & Inequalities

8 methods · 9 questions

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To establish an inequality, rearrange it so one side minus the other is a single non-negative object: a sum of squares, a square completed in one variable, or the Cauchy-Schwarz defect; equality then corresponds exactly to that object vanishing.

Trigger: a $⩾$ / $⩽$ to prove for all real values, or a request for the equality condition.

Instances:
(i) Write the difference as an explicit sum of squares so non-negativity is manifest.
(ii) Recognise $(a x + b y + cz)^{2} ⩽ (a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2})$ as Cauchy-Schwarz via a scalar product, equality iff the vectors are proportional.
(iii) Translate a geometric extremal claim (max area, longest diagonal) into an algebraic square-completion or scalar-product bound.
(iv) Chain elementary integer inequalities $b c ⩾ 2 c ⩾ 2 + c$ step by step, each step a small non-negative difference.

Linked questions (2)

STEP 2 2006 · Q6

By considering a suitable scalar product, prove that

(a x + b y + cz)^{2} ⩽ (a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2})

for any real numbers

a

b

c

x

y

and

z

. Deduce a necessary and sufficient condition on

a

b

c

x

y

and

z

for the following equation to hold:

(a x + b y + cz)^{2} = (a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) .

Show that $(x + 2 y + 2 z)^{2} ⩽ 9 (x^{2} + y^{2} + z^{2})$ for all real numbers $x$ , $y$ and $z$ .
Find real numbers $p$ , $q$ and $r$ that satisfy both $p^{2} + 4 q^{2} + 9 r^{2} = 729 and 8 p + 8 q + 3 r = 243 .$

Worked solution

The key idea is to interpret the left-hand side as a dot product. Define vectors

u = (a, b, c)

and

v = (x, y, z)

. Their dot product is

u \cdot v = a x + b y + cz

, and the standard formula gives

u \cdot v = ∣ u ∣∣ v ∣ cos θ

, where

θ

is the angle between the two vectors.

Squaring both sides:

(a x + b y + cz)^{2} = ∣ u ∣^{2} ∣ v ∣^{2} cos^{2} θ = (a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) cos^{2} θ .

Since

cos^{2} θ ⩽ 1

for all real

θ

, dividing through gives the required inequality

(a x + b y + cz)^{2} ⩽ (a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) .

Equality holds precisely when

cos^{2} θ = 1

, that is, when

θ = 0

θ = π

. Geometrically, this means

u

and

v

are parallel: one is a scalar multiple of the other. So the necessary and sufficient condition is that

(x, y, z) = λ (a, b, c)

for some real

λ

(or that one of the vectors is zero).

Part (i). Take

a = 1

b = 2

c = 2

, so

a^{2} + b^{2} + c^{2} = 1 + 4 + 4 = 9

. The Cauchy–Schwarz inequality immediately yields

(x + 2 y + 2 z)^{2} ⩽ 9 (x^{2} + y^{2} + z^{2})

for all real

x, y, z

.

For completeness, equality holds when

(x, y, z) = λ (1, 2, 2)

, i.e.

y = 2 x /1

— but the question only asks for the inequality.

Part (ii). Observe that

p^{2} + 4 q^{2} + 9 r^{2}

can be written as

p^{2} + (2 q)^{2} + (3 r)^{2}

, and

8 p + 8 q + 3 r

pairs naturally with vectors

u = (p, 2 q, 3 r)

and

v = (8, 4, 1)

.

Compute

v

: to match the form

a x + b y + cz

with

(p, 2 q, 3 r)

, write

8 p + 8 q + 3 r = 8 p + 4 (2 q) + 1 (3 r)

. So take

u = (p, 2 q, 3 r)

and

v = (8, 4, 1)

.

Then

∣ u ∣^{2} = p^{2} + 4 q^{2} + 9 r^{2} = 729

and

∣ v ∣^{2} = 64 + 16 + 1 = 81

, giving

∣ u ∣ = 27

and

∣ v ∣ = 9

.

Cauchy–Schwarz gives

(8 p + 4 (2 q) + (3 r))^{2} ⩽ 729 \times 81

, so

8 p + 8 q + 3 r ⩽ 243

. The second condition states

8 p + 8 q + 3 r = 243

, which is the equality case. Hence

(p, 2 q, 3 r) = λ (8, 4, 1)

for some

λ

, meaning

p = 8 λ

q = 2 λ

r = \frac{λ}{3}

.

Substituting into

p^{2} + 4 q^{2} + 9 r^{2} = 729

64 λ^{2} + 16 λ^{2} + λ^{2} = 81 λ^{2} = 729

, so

λ^{2} = 9

and

λ = 3

(taking the positive root to satisfy

8 p + 8 q + 3 r = 243 > 0

).

Therefore

p = 24

q = 6

r = 1

STEP 2 2022 · Q5

5 (i) Given that

a > b > c > 0

are constants, and that

x, y, z

are non-negative variables, show that

a x + b y + cz \leq a (x + y + z) .

In the acute-angled triangle

A B C

a

b

and

c

are the lengths of sides

B C

C A

and

A B

, respectively, with

a > b > c

P

is a point inside, or on the sides of, the triangle, and

x

y

and

z

are the perpendicular distances from

P

B C

C A

and

A B

, respectively. The area of the triangle is

Δ

.

(ii) (a) Find

Δ

in terms of

a

b

c

x

y

and

z

.

(b) Find both the minimum value of the sum of the perpendicular distances from

P

to the three sides of the triangle and the values of

x

y

and

z

which give this minimum sum, expressing your answers in terms of some or all of

a

b

c

and

Δ

.

(iii) (a) Show that, for all real

a

b

c

x

y

and

z

(a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) = (b x - a y)^{2} + (cy - b z)^{2} + (a z - c x)^{2} + (a x + b y + cz)^{2} .

(b) Find both the minimum value of the sum of the squares of the perpendicular distances from

P

to the three sides of the triangle and the values of

x

y

and

z

which give this minimum sum, expressing your answers in terms of some or all of

a

b

c

and

Δ

.

(iv) Find both the maximum value of the sum of the squares of the perpendicular distances from

P

to the three sides of the triangle and the values of

x

y

and

z

which give this maximum sum, expressing your answers in terms of some or all of

a

b

c

and

Δ

Worked solution

Part (i). Since

b < a

and

y \geq 0

, we have

b y \leq a y

. Similarly, since

c < a

and

z \geq 0

, we have

cz \leq a z

. Adding

a x

to both sides of the combined inequality:

a x + b y + cz \leq a x + a y + a z = a (x + y + z) .

Equality holds when

y = z = 0

.

Part (ii)(a). Draw lines from

P

to each vertex, splitting triangle

A B C

into three smaller triangles:

P B C

P C A

, and

P A B

, with bases

a, b, c

and corresponding heights

x, y, z

. Summing their areas:

Δ = \frac{1}{2} a x + \frac{1}{2} b y + \frac{1}{2} cz .

Part (ii)(b). The constraint from (ii)(a) is

a x + b y + cz = 2Δ

. Applying part (i):

2Δ = a x + b y + cz \leq a (x + y + z) ⟹ x + y + z \geq \frac{2Δ}{a} .

The minimum value of

x + y + z

\frac{2Δ}{a}

, achieved at

x = \frac{2Δ}{a}

y = 0

z = 0

. Geometrically,

P

lies on side

B C

(the longest side).

Part (iii)(a). Expand the right-hand side term by term:

(b x - a y)^{2} = b^{2} x^{2} - 2 ab x y + a^{2} y^{2},

(cy - b z)^{2} = c^{2} y^{2} - 2 b cy z + b^{2} z^{2},

(a z - c x)^{2} = a^{2} z^{2} - 2 a c x z + c^{2} x^{2},

(a x + b y + cz)^{2} = a^{2} x^{2} + b^{2} y^{2} + c^{2} z^{2} + 2 ab x y + 2 b cy z + 2 a c x z .

Summing all four expressions, every cross term cancels in pairs, leaving:

(a^{2} + b^{2} + c^{2}) x^{2} + (a^{2} + b^{2} + c^{2}) y^{2} + (a^{2} + b^{2} + c^{2}) z^{2} = (a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) .

Part (iii)(b). Substituting

a x + b y + cz = 2Δ

into the identity from (iii)(a):

(a^{2} + b^{2} + c^{2}) (x^{2} + y^{2} + z^{2}) = (b x - a y)^{2} + (cy - b z)^{2} + (a z - c x)^{2} + 4 Δ^{2} \geq 4 Δ^{2} .

Hence

x^{2} + y^{2} + z^{2} \geq \frac{4 Δ ^{2}}{a ^{2} + b ^{2} + c ^{2}}

, with equality when

b x = a y

cy = b z

a z = c x

, i.e. when

\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k

.

Substituting into the area constraint:

\frac{1}{2} k (a^{2} + b^{2} + c^{2}) = Δ

, so

k = \frac{2Δ}{a ^{2} + b ^{2} + c ^{2}}

. The minimum is attained at:

x = \frac{2 a Δ}{a ^{2} + b ^{2} + c ^{2}}, y = \frac{2 b Δ}{a ^{2} + b ^{2} + c ^{2}}, z = \frac{2 c Δ}{a ^{2} + b ^{2} + c ^{2}} .

Part (iv). Since

c < b < a

, the inequality reverses:

a x + b y + cz \geq c (x + y + z)

. Squaring (both sides non-negative):

(a x + b y + cz)^{2} \geq c^{2} (x + y + z)^{2} .

For non-negative

x, y, z

we also have

(x + y + z)^{2} \geq x^{2} + y^{2} + z^{2}

(since cross terms are non-negative). Combining:

c^{2} (x^{2} + y^{2} + z^{2}) \leq (a x + b y + cz)^{2} = 4 Δ^{2} ⟹ x^{2} + y^{2} + z^{2} \leq \frac{4 Δ ^{2}}{c ^{2}} .

The maximum value is

\frac{4 Δ ^{2}}{c ^{2}}

, achieved when

x = 0

y = 0

z = \frac{2Δ}{c}

. Here

P

lies on side

A B

(the shortest side), with the entire "area contribution" coming from the perpendicular to

A B