Prove n2>(4n+7) for n≥6 w/ Induction

[mamma_mia66]

Homework Statement

Prove by induction that n²>(4n+7) for all integers n$$\geq$$6.
[Hint: somewhere you may use the fact that (2k+1)$$\geq$$13 > 4.]

Homework Equations

The Attempt at a Solution

n²> (4n+7)

(B) when n =6
6²> (4*6+7)
36>31 true

(I) given n=k then k²> (4k+7)

let n=k+1= k²+2k+1

(k²+ 2k+1) > 4k+7 +2k+1

4k+7+2k+1> 4k+7+4

6k+8 > 4(k+1)+7

Is that looks like okay? Please help. Thank you.

 

[Mark44]

It's hard to follow the logic in what you wrote.

For the step you're trying to prove, you should have something like this:
(k + 1)^2 > ... > 4(k + 1) + 7
In other words, your inequality starts off with (k + 1)^2 and ends with a smaller quantity, 4(k + 1) + 7.

 

[mamma_mia66]

Okay. Thank you.

 

[HallsofIvy]

Homework Statement

Prove by induction that n²>(4n+7) for all integers n$$\geq$$6.
[Hint: somewhere you may use the fact that (2k+1)$$\geq$$13 > 4.]

Homework Equations

The Attempt at a Solution

n²> (4n+7)

(B) when n =6
6²> (4*6+7)
36>31 true

(I) given n=k then k²> (4k+7)

let n=k+1= k²+2k+1

Surely this is not what you meant to write! you meant, I think, that "let n= k+1, then n²= (k+1)²= k²+ 2k+ 1

[/quote](k²+ 2k+1) > 4k+7 +2k+1

4k+7+2k+1> 4k+7+4 [/quote]
It would be useful to point out that because k> 5, 2k+1> 11> 4

6k+8 > 4(k+1)+7

Is that looks like okay? Please help. Thank you.