Algebra Theory - Inequalities

1. Basic operations on inequalities:
Given a > b

a + c > b + c
a - c > b - c
a c > b c if c > 0
a c < b c if c < 0
a/c > b/c if c > 0
a/c < b/c if c < 0
a² > b² if |a| > |b| (Even power)
a² < b² if |a| < |b|
a³ > b³ (Odd power)

If a ≥ b and c ≥ d:

a + c ≥ b + d
a c ≥ b d if a, b, c, d > 0

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2. Modulus inequality

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}

→ $- ∣ x ∣ \leq x \leq ∣ x ∣$

→ $|a+b| \leq |a| + |b|$

$∣ a_{1} + a_{2} + a_{3} + \dots + a_{n} ∣ \leq ∣ a_{1} ∣ + ∣ a_{2} ∣ + \dots + ∣ a_{n} ∣$

→ $| |a| - |b| | \leq |a-b|$

$|a+b| \leq |a| + |b|$

\begin{align*} |a+b| = |a| + |b| &\qquad ab > 0 \\ |a+b| < |a| + |b| &\qquad ab < 0 \end{align*}

Triangle Inequality

\Rightarrow \quad \underline{a + b > c} \\ \quad b + c > a \\ \quad c + a > b

\Rightarrow |a - c| < b \\ \quad |b - c| < a \\ \quad |a - b| < c

|a - b| < c < a + b

\begin{cases} a + b - c > 0 \\ b + c - a > 0 \\ c + a - b > 0 \end{cases} \Rightarrow (a + b - c)(b + c - a)(c + a - b) > 0

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Sum of Square Inequality

\sum (x^2) \geq 0

Ex1. Prove that:

x^2 + y^2 + z^2 \geq xy + yz + zx

Proof:

x^2 + y^2 + z^2 - xy - yz - zx \geq 0

2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx \geq 0

x^2 - 2xy + y^2 + y^2 - 2yz + z^2 + z^2 - 2zx + x^2 \geq 0

(x - y)^2 + (y - z)^2 + (z - x)^2 \geq 0

(\cdot)^2 + (\cdot)^2 + (\cdot)^2 \geq 0

This uses the sum of squares identity, where the sum of non-negative squares is always non-negative, completing the proof.

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Ex2. Prove that:

2016x^2 + 2016y^2 + 6z^2 \geq 2(2013xy + 3yz + 3zx)

Steps:

2013x^2 + 2013y^2 - 2(2013xy) \geq 0

3x^2 + 3z^2 - 2(3zx) \geq 0

3y^2 + 3z^2 - 2(3yz) \geq 0

Adding the three inequalities:

2013(x - y)^2 + 3(x - z)^2 + 3(y - z)^2 \geq 0 \quad \checkmark

Ex3. Prove that:

a^2 + 5b^2 + 4c^2 \geq 4ab + 4bc

Ex4. Prove that (if $x, y, z > 0$ ):

\frac{x^2 + yz}{y + z} + \frac{y^2 + zx}{z + x} + \frac{z^2 + xy}{x + y} \geq x + y + z

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3.

$a^2 + 4b^2 - 4ab + b^2 + 4c^2 - 4bc \geq 0$ $(a - 2b)^2 + (b - 2c)^2 \geq 0$

4.

Prove:

$\frac{x^2 + yz}{y + z} + \frac{y^2 + zx}{z + x} + \frac{z^2 + xy}{x + y} \geq x + y + z$

Steps:

Start with the expression:

$\frac{x^2 + yz}{y + z} - x + \frac{y^2 + zx}{z + x} - y + \frac{z^2 + xy}{x + y} - z \geq 0$

Now simplify each term:

$\frac{x^2 + yz - yx - xz}{y + z} + \frac{y^2 + zx - zx - yz}{z + x} + \frac{z^2 + xy - xy - zx}{x + y}$ $= \frac{(x - y)(x - z)}{y + z} + \frac{(y - z)(y - x)}{z + x} + \frac{(z - x)(z - y)}{x + y} \geq 0$

Multiply out the squares and rearrange:

$\frac{(x^2 - yz)(x^2 - z^2) + (y^2 - z^2)(y^2 - x^2) + (z^2 - x^2)(z^2 - y^2)}{(x + y)(y + z)(z + x)} \geq 0$

Final step:

$\frac{x^4 + y^4 + z^4 - x^2 y^2 - y^2 z^2 - z^2 x^2}{(x + y)(y + z)(z + x)} \geq 0$

$\frac{(x^2 - y^2)^2 + (y^2 - z^2)^2 + (z^2 - x^2)^2}{(x + y)(y + z)(z + x)} \geq 0$ $\Rightarrow \text{By sum of squares: } N^r \geq 0, \quad D^r > 0 \quad \text{(and as per question)}$ $\Rightarrow \frac{N^r}{D^r} \geq 0$

Where $N^r$ represents the numerator and $D^r$ the denominator. Since the numerator is a sum of squares (always $\geq 0$ ) and the denominator is positive (due to $x, y, z > 0$ ), the entire expression is non-negative.

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Quadratic Inequality

Let:

$f(x) = ax^2 + bx + c \quad \text{and} \quad b^2 - 4ac < 0$

We want to solve:

$ax^2 + bx + c > 0$

Example:

$y = x^2 + 3x + 2 = (x + 1)(x + 2)$

Sign analysis:

$x < -2$ : $(−)(−) \rightarrow +$
$-2 < x < -1$ : $(+)(−) \rightarrow -$
$x > -1$ : $(+)(+) \rightarrow +$

General Case:

$y = (x - \alpha)(x - \beta), \quad \alpha < \beta$

Sign table:

$x > \beta$ : $(+)(+) \rightarrow +$
$\alpha < x < \beta$ : $(+)(−) \rightarrow -$ ✅
$x < \alpha$ : $(−)(−) \rightarrow +$

Conclusion:

If $a > 0$ , then:

$ax^2 + bx + c \text{ will be negative between the roots.}$

Also:

If $b^2 - 4ac < 0$ , then no real roots, and:
$ax^2 + bx + c > 0$
If $b^2 - 4ac \leq 0$ , and $a > 0$ , then:
$ax^2 + bx + c \geq 0$

(1) Prove that:

$a^2 + ac + c^2 \geq 3b(a - b + c)$

Rewriting the inequality:

$a^2 + ac + c^2 - 3ab + 3b^2 - 3bc \geq 0$

Group terms:

$3b^2 - 3(a + c)b + a^2 + ac + c^2 \geq 0$

This is a quadratic in $b$ :

$3b^2 - 3(a + c)b + (a^2 + ac + c^2) \geq 0$

Discriminant approach:

Let:

$D = 9(a + c)^2 - 4(3)(a^2 + ac + c^2)$ $= 9a^2 + 9c^2 + 18ac - 12a^2 - 12ac - 12c^2$ $= -3a^2 - 3c^2 + 6ac = -3(a - c)^2$

Since $(a - c)^2 \geq 0$ , we have:

$-3(a - c)^2 \leq 0 \Rightarrow D \leq 0$

Conclusion:

$3b^2 - 3(a + c)b + a^2 + ac + c^2 \geq 0$

Thus, the inequality is proven.

AM ≥ GM ≥ HM Inequality

Definitions:

Let $a_1, a_2, \ldots, a_n$ be positive real numbers.

Arithmetic Mean (AM):
$\text{AM} = \frac{a_1 + a_2 + a_3 + \cdots + a_n}{n}$
Geometric Mean (GM):
$\text{GM} = \left( a_1 a_2 a_3 \cdots a_n \right)^{1/n}$
Harmonic Mean (HM):
$\text{HM} = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \cdots + \frac{1}{a_n}}$

For two numbers $a$ and $b$ :

$\text{AM} = \frac{a + b}{2}$
$\text{GM} = \sqrt{ab}$
$\text{HM} = \frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a + b}$

Inequality Summary:

If $a_1, a_2, \ldots, a_n$ are all positive numbers, then:

$\text{AM} \geq \text{GM} \geq \text{HM}$

Equality holds if and only if:

$a_1 = a_2 = \cdots = a_n = k \quad \Rightarrow \quad \text{AM} = \text{GM} = \text{HM} = k$

Example: Numbers 1, 2, 4

Arithmetic Mean (AM):

$\text{AM}(1, 2, 4) = \frac{1 + 2 + 4}{3} = \frac{7}{3} \approx 2.33$

Geometric Mean (GM):

$\text{GM}(1, 2, 4) = (1 \cdot 2 \cdot 4)^{1/3} = 8^{1/3} = 2$

Harmonic Mean (HM):

$\text{HM}(1, 2, 4) = \frac{3}{\frac{1}{1} + \frac{1}{2} + \frac{1}{4}} = \frac{3}{1 + 0.5 + 0.25} = \frac{3}{1.75} = \frac{12}{7} \approx 1.71$

Proof of AM ≥ GM

Let $a$ , $b$ be positive real numbers.

$(\sqrt{a} - \sqrt{b})^2 \geq 0$ $a + b - 2\sqrt{ab} \geq 0 \Rightarrow a + b \geq 2\sqrt{ab}$

Divide both sides by 2:

$\frac{a + b}{2} \geq \sqrt{ab}$

Hence:

$\text{AM} \geq \text{GM}$

(1) Prove that:

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 \quad \text{where } a, b, c > 0$

Using the AM ≥ GM inequality on

$\frac{a}{b}, \quad \frac{b}{c}, \quad \frac{c}{a}$
By the AM-GM inequality:
$\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq \left( \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} \right)^{1/3}$
Simplify the right-hand side:
$\left( \frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a} \right)^{1/3} = (1)^{1/3} = 1$
So:
$\frac{\frac{a}{b} + \frac{b}{c} + \frac{c}{a}}{3} \geq 1 \Rightarrow \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3$

✅ Hence, the inequality is proven using AM ≥ GM for positive real numbers.

(2) Given:

Let $a, b, c, d$ be positive real numbers such that:
$abcd = 1$
Then prove that:
$(1 + a)(1 + b)(1 + c)(1 + d) \geq 16$

Proof using AM ≥ GM:

Step 1:

Apply AM ≥ GM on the pair $(1, a)$ :
$\frac{1 + a}{2} \geq \sqrt{a} \Rightarrow 1 + a \geq 2\sqrt{a} \quad \text{(✓)}$
Similarly:

$1 + b \geq 2\sqrt{b} \quad \text{(✓)}$

$1 + c \geq 2\sqrt{c} \quad \text{(✓)}$

$1 + d \geq 2\sqrt{d} \quad \text{(✓)}$

Step 2:

Multiply all four inequalities:
$(1 + a)(1 + b)(1 + c)(1 + d) \geq 2^4 \sqrt{abcd}$
Since $abcd = 1$ , we have $\sqrt{abcd} = \sqrt{1} = 1$
So:
$(1 + a)(1 + b)(1 + c)(1 + d) \geq 16$
✅ Hence proved.

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(3) Let $a_1, a_2, \ldots, a_n$ be positive numbers.

Prove that:

$\sqrt{a_1 a_2} + \sqrt{a_1 a_3} + \cdots + \sqrt{a_1 a_n} + \sqrt{a_2 a_3} + \sqrt{a_2 a_4} + \cdots + \sqrt{a_2 a_n} + \cdots + \sqrt{a_{n-1} a_n}$ $\leq \frac{n - 1}{2} (a_1 + a_2 + a_3 + \cdots + a_n)$

Proof using AM ≥ GM:

From AM ≥ GM:
$\frac{a_1 + a_2}{2} \geq \sqrt{a_1 a_2} \quad \text{✓}$ $\frac{a_1 + a_3}{2} \geq \sqrt{a_1 a_3} \quad \text{✓}$ $\vdots$ $\frac{a_{n-1} + a_n}{2} \geq \sqrt{a_{n-1} a_n} \quad \text{✓}$

Adding all these inequalities:

Each pair $(a_i + a_j)/2$ appears once for each square root term, and so:
$\frac{(a_1 + a_2) + (a_1 + a_3) + \cdots + (a_1 + a_n) + (a_2 + a_3) + \cdots + (a_{n-1} + a_n)}{2} \geq \sum_{1 \le i < j \le n} \sqrt{a_i a_j}$
There are $\frac{n(n-1)}{2}$ such square root terms on the RHS, and each $a_i$ appears $n - 1$ times in the sum on the LHS.
So:
$\frac{(n-1)(a_1 + a_2 + \cdots + a_n)}{2} \geq \sum_{1 \le i < j \le n} \sqrt{a_i a_j}$
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(IV)

If $a + b + c = 3$ , where $a, b, c > 0$ ,

Find:

(a) The maximum value of $abc$
(b) The maximum value of $a^2 b^3 c$

(a) Max value of $abc$

By AM ≥ GM for $a, b, c$ :
$\frac{a + b + c}{3} \geq (abc)^{1/3} \Rightarrow \frac{3}{3} \geq (abc)^{1/3} \Rightarrow 1 \geq (abc)^{1/3} \Rightarrow abc \leq 1$
✅ So, the maximum value of $abc$ is:
$\boxed{1}$

(b) Max value of $a^2 b^3 c$

We are given:
$a + b + c = 3$
Write the sum as:
$\frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + c = 3$
Apply AM ≥ GM on the 6 terms:
$\left\{ \frac{a}{2}, \frac{a}{2}, \frac{b}{3}, \frac{b}{3}, \frac{b}{3}, c \right\}$ $\frac{\frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{b}{3} + c}{6} \geq \left( \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{b}{3} \cdot \frac{b}{3} \cdot \frac{b}{3} \cdot c \right)^{1/6}$ $\frac{a + b + c}{6} \geq \left( \frac{a^2 b^3 c}{2^2 \cdot 3^3} \right)^{1/6}$ $\frac{3}{6} \geq \left( \frac{a^2 b^3 c}{2^2 \cdot 3^3} \right)^{1/6} \Rightarrow \left( \frac{1}{2} \right)^6 \geq \frac{a^2 b^3 c}{2^2 \cdot 3^3}$ $\Rightarrow a^2 b^3 c \leq \frac{3^3}{2^2} = \frac{27}{4}$
⚠ Small correction:
The denominator should be:
$2^2 \cdot 3^3 = 4 \cdot 27 = 108 \Rightarrow a^2 b^3 c \leq \frac{27}{16}$
✅ So, the maximum value of $a^2 b^3 c$ is:
$\boxed{\frac{27}{16}}$