Algebra theory inequalities continued

Ex1:
$x^4 + px^3 + qx^2 + rx + s = 0$ has 4 positive real roots. Then prove that

a) $pr - 16s \geq 0$
b) $q^2 - 36s \geq 0$

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Let $\alpha, \beta, \gamma, \delta$ be the roots.

$$\begin{aligned} S_1 &= \alpha + \beta + \gamma + \delta = -p \\ S_2 &= \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = q \\ S_3 &= \alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = -r \\ S_4 &= \alpha\beta\gamma\delta = s \end{aligned}$$

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(a) $pr \geq 16s$

Using AM ≥ GM for $\{ \alpha, \beta, \gamma, \delta \}$:

$$\frac{\alpha + \beta + \gamma + \delta}{4} \geq (\alpha \beta \gamma \delta)^{1/4} \Rightarrow \frac{-p}{4} \geq s^{1/4}$$

Using AM ≥ GM for $\{ \alpha\beta\gamma, \alpha\beta\delta, \alpha\gamma\delta, \beta\gamma\delta \}$:

$$\frac{\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta}{4} \geq (\alpha^3 \beta^3 \gamma^3 \delta^3)^{1/4} \Rightarrow \frac{-r}{4} \geq s^{3/4}$$

Multiplying the two inequalities:

$$\frac{pr}{16} \geq s \Rightarrow pr \geq 16s$$

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(b) $q^2 \geq 36s$

Using AM ≥ GM for $\{ \alpha\beta, \alpha\gamma, \alpha\delta, \beta\gamma, \beta\delta, \gamma\delta \}$:

$$\frac{\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta}{6} \geq (\alpha^2 \beta^2 \gamma^2 \delta^2)^{1/6} \Rightarrow \frac{q}{6} \geq s^{1/2} \Rightarrow q^2 \geq 36s$$

Ex2: Prove that if a,b,c are positive numbers then:

$$\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c} \geq 6$$

Method1:
Add 1 to each fraction:
(b+c)/a + 1 + ... >= 9
=> (a+b+c)[1/a + 1/b + 1/c] >= 9
=> (a+b+c)/3 >= 3/[1/a + 1/b + 1/c]
Which is nothing but 
A.M. >= H.M. for a,b,c
H.P.

Method2:
Rewrite the given inequality as:
b/a + c/a + c/b + a/b + a/c + b/c >= 6
[b/a + c/a + c/b + a/b + a/c + b/c]/6 >= 1

1 on the R.H.S can be written as:
[b/a * c/a * c/b * a/b * a/c * b/c] since everything cancels out.
So again it's proven using A.M. >= G.M.

Ex3:

Given $a > 0$, prove that:

$$\left( a^3 + a^2 + a + 1 \right)^2 \geq 16a^3$$

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Solution:
Simply do A.M. >= G.M. for the given 4 numbers
(1 + a + a^2 + a^3)/4 >= (1 * a * a^2 * a^3)^(1/4) = a^(3/2)
=> (1 + a + a^2 + a^3) >= = 4.a^(3/2)
=> (1 + a + a^2 + a^3)^2 >= = 16.a^(3)
H.P.

Ex4.

$a_1, a_2, a_3, \ldots, a_n$ are positive numbers.

Prove that

$$n(a_1 a_2 a_3 \cdots a_n) \leq a_1^n + a_2^n + \cdots + a_n^n$$

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Solution:
Straightforward application of A.M. >= G.M. on a1^n, a2^n ... an^n

Ex5.

Given:

$$0 < a, b, c < 1 \quad \text{and} \quad a + b + c = 2$$

Prove that:

$$\frac{a}{1 - a} \cdot \frac{b}{1 - b} \cdot \frac{c}{1 - c} \geq 8$$

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Solution:

Given:

$$a + b + c = 2 \quad \text{and} \quad 0 < a, b, c < 1$$

Rewrite the given inequality as:

$$a \cdot b \cdot c \geq 2(1 - a) \cdot 2(1 - b) \cdot 2(1 - c)$$ $$abc \geq (2 - 2a)(2 - 2b)(2 - 2c)$$ $$abc \geq (b + c - a)(a + c - b)(a + b - c)$$

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Using AM ≥ GM:

For the pair $\{a + c - b, a + b - c\}$:

$$\frac{a + c - b + a + b - c}{2} \geq \sqrt{(a + c - b)(a + b - c)} \Rightarrow a \geq \sqrt{(a + c - b)(a + b - c)} \quad \text{(i)}$$

For $\{b + c - a, b + a - c\}$:

$$\frac{b + c - a + b + a - c}{2} \geq \sqrt{(b + c - a)(b + a - c)} \Rightarrow b \geq \sqrt{(b + c - a)(b + a - c)} \quad \text{(ii)}$$

For $\{c + a - b, c + b - a\}$:

$$\Rightarrow c \geq \sqrt{(c + a - b)(c + b - a)} \quad \text{(iii)}$$

Multiplying (i), (ii), and (iii):

$$abc \geq (a + c - b)(b + c - a)(a + b - c)$$

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Ex6.

$a, b, c$ are positive numbers such that

$$a + b + c = 1$$

Then prove that:

$$(1 + a)(1 + b)(1 + c) \geq 8(1 - a)(1 - b)(1 - c)$$

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From the condition:

$$a + b + c = 1 \Rightarrow 1 + a + b + c = 2 \Rightarrow 1 + a = 2 - b - c \Rightarrow 1 + a = (1 - b) + (1 - c)$$

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Using AM ≥ GM for $\{1 - b, 1 - c\}$:

$$\frac{1 - b + 1 - c}{2} \geq \sqrt{(1 - b)(1 - c)} \Rightarrow 1 + a \geq 2\sqrt{(1 - b)(1 - c)}$$

Similarly:

$$1 + b \geq 2\sqrt{(1 - c)(1 - a)} \\ 1 + c \geq 2\sqrt{(1 - a)(1 - b)}$$

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Multiplying all three inequalities:

$$(1 + a)(1 + b)(1 + c) \geq 8 \sqrt{(1 - b)(1 - c)(1 - c)(1 - a)(1 - a)(1 - b)} \Rightarrow (1 + a)(1 + b)(1 + c) \geq 8(1 - a)(1 - b)(1 - c)$$

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Ex7:

$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1, \quad \text{with } x, y, z > 0$$

Prove that:

$$(x - 1)(y - 1)(z - 1) \geq 8$$

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Approach using AM ≥ GM on pairs:

$$\frac{1}{x} + \frac{1}{y} \geq 2\sqrt{\frac{1}{xy}} \\ \frac{1}{y} + \frac{1}{z} \geq 2\sqrt{\frac{1}{yz}} \\ \frac{1}{z} + \frac{1}{x} \geq 2\sqrt{\frac{1}{zx}}$$

So, in general:

$$\left( \frac{1}{x} + \frac{1}{y} \right)\left( \frac{1}{y} + \frac{1}{z} \right)\left( \frac{1}{z} + \frac{1}{x} \right) \geq \frac{8}{xyz}$$

Now take:

$$\left(1 - \frac{1}{x}\right)\left(1 - \frac{1}{y}\right)\left(1 - \frac{1}{z}\right) = (x - 1)(y - 1)(z - 1) \cdot \frac{1}{xyz}$$

So multiplying both sides:

$$(x - 1)(y - 1)(z - 1) \geq 8$$

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Ex8:

Prove that:

$$a^8(1 + b^8) + b^8(1 + c^8) + c^8(1 + d^8) + d^8(1 + a^8) > 8a^2b^3c^2d^3$$

Given that $a, b, c, d$ are 4 distinct positive numbers.

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Solution:

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Given:

$$a^8(1 + b^8) + b^8(1 + c^8) + c^8(1 + d^8) + d^8(1 + a^8)$$

Expanded as:

$$a^8 + a^8b^8 + b^8 + b^8c^8 + c^8 + c^8d^8 + d^8 + d^8a^8$$

Apply AM ≥ GM on these 8 terms:

$$\frac{a^8 + a^8b^8 + b^8 + b^8c^8 + c^8 + c^8d^8 + d^8 + d^8a^8}{8} \geq \left( a^{24} b^{24} c^{24} d^{24} \right)^{\frac{1}{8}}$$

So:

$$a^8 + a^8b^8 + b^8 + b^8c^8 + c^8 + c^8d^8 + d^8 + d^8a^8 \geq 8a^3b^3c^3d^3$$

Thus:

$$a^8(1 + b^8) + b^8(1 + c^8) + c^8(1 + d^8) + d^8(1 + a^8) \geq 8a^3b^3c^3d^3$$

Equality in AM = GM holds only if $a = b = c = d$, but since the values are 4 distinct positive numbers, strict inequality holds:

$$a^8(1 + b^8) + b^8(1 + c^8) + c^8(1 + d^8) + d^8(1 + a^8) > 8a^3b^3c^3d^3$$

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Ex9:

Let $S = a + b + c + d$, where $a, b, c, d$ are 4 positive numbers.

Prove that:

$$(S - a)(S - b)(S - c)(S - d) \geq 81 \cdot abcd$$

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Solution:

Prove that:

$$(S - a)(S - b)(S - c)(S - d) \geq 81 \cdot abcd$$

Where $S = a + b + c + d$

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From the definition:

$$S - a = b + c + d$$

Apply AM ≥ GM to $\{b, c, d\}$:

$$\frac{b + c + d}{3} \geq (bcd)^{1/3} \Rightarrow b + c + d \geq 3(bcd)^{1/3} \Rightarrow S - a \geq 3(bcd)^{1/3}$$

Similarly:

$$S - b \geq 3(acd)^{1/3} \\ S - c \geq 3(abd)^{1/3} \\ S - d \geq 3(abc)^{1/3}$$

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Multiply all four inequalities:

$$(S - a)(S - b)(S - c)(S - d) \geq 3^4 \cdot (bcd \cdot acd \cdot abd \cdot abc)^{1/3}$$

Simplifying the product inside the cube root:

$$= 81 \cdot (a^3 b^3 c^3 d^3)^{1/3} = 81 \cdot abcd$$

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Thus, the inequality is proven:

$$(S - a)(S - b)(S - c)(S - d) \geq 81 \cdot abcd$$