I need help!! I have NO idea how to prove. Please help. I've been doing this for an hour already :(

Let N e R (Let any real number be..)

For n≥3, prove that 2n^2≥ (n+1)^2

For n>2, prove that (3n)/(n^2 +2) <1

For n≥4, 2n^3 ≥ (n+1)^3

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I need help!! I have NO idea how to prove. Please help. I've been doing this for an hour already :(

Let N e R (Let any real number be..)

For n≥3, prove that 2n^2≥ (n+1)^2

For n>2, prove that (3n)/(n^2 +2) <1

For n≥4, 2n^3 ≥ (n+1)^3

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## Answers (2)

1.

Prove 2n² ≥ (n+1)² for n≥3

2n² ≥ n² + 2n + 1

n² - 2n ≥ 1

n(n-2) ≥ 1 ..............have to prove this for n≥3

if n=3 then have n(n-2) = 3(1) = 3

so if n>3 then n(n-2)> 3(1) > 3 hence

n(n-2) ≥1 when n≥3 so

2n² ≥ n+1)² when n≥3

2.

prove (3n)/(n²+2) < 1 for n>2

(3n)/(n²+2) < 1

3n < n² + 2 ...............(n²+2) is always positive for all n so inequality sign is unchanged

n² - 3n + 2 > 0

(n-1)(n-2) > 0 ...........have to prove this for n>2

either (n-1) AND (n-2) are both positive OR both negative

if n>2 then (n-1) > 1 (ie: positive) AND (n-2) > 0 (ie: positive)

Hence,

(n-1)(n-2) > 0 for n>2 so

(3n)/(n²+2) < 1 for n > 2

3.

prove 2n³ ≥ (n+1)³ for n ≥ 4

2n³ ≥ (n+1)³

2n³ ≥ n³ + 3n² + 3n + 1

n³ - 3n² - 3n ≥ 1

n(n² - 3n - 3) ≥ 1 ..............have to prove this for n ≥ 4

(n² - 3n - 3) is a quadratic. Its sketch (y vs n) would be an upright parabola with n-intercepts of ≈ -0.79 and ≈ 3.79 (used quadratic formula). You could do a quick sketch here (pen and paper comes in handy)

So for n ≥ 4 we know (from sketch) that (n² - 3n - 3) ≥ 0 (bcoz graph is above horizontal axis)

In fact if n=4 => (n² - 3n - 3) = 1

Hence:

if n = 4 then n(n² - 3n - 3) = 4(1) So

if n ≥ 4 then (n² - 3n - 3) ≥ 1 as the graph is increasing over that interval and

n(n² - 3n - 3) ≥ 4(1) ≥ 4 so

n(n² - 3n - 3) ≥ 1 for n ≥ 4 hence

2n³ ≥ (n+1)³ for n ≥ 4

Just plug it in plug it in