Thread: Inequalities

Can anyone help me confirm if I have solved this correctly?

Many thanks.

Q. (1 + 2x)ⁿ > 1 + 2nx, for x > 0 & n

Attempt: Step 1: For n = 1...
(1 + 2x)¹ = 1 + 2x '&' 1 + 2(1)x = 1 + 2x
Since 1 + 2x > 1 + 2x, n is true.

Step 2: Assume the statement is true for n = k, i.e. assume (1 + 2x)^k > 1 + 2kx,
We must now show that the statement is true for n = k + 1,
i.e. (1 + 2x)^k+1 > 1 + 2(k + 1)x = 1 + 2kx + 2x
(1 + 2x)^k+1 = (1 + 2x)¹(1 + 2x)^k > (1 + 2x)(1 + 2kx)...((1 + 2x)^k > 1 + 2kx...assumed)
If (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x then (1 + 2x)^k+1 > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) - (1 + 2kx + 2x) > 0
if 2kx² + 2kx + 2x + 1 - 1 - 2kx - 2x > 0
if 2kx² > 0
if x²[2k] > 0...true when x > 0 & n > 1
(1 + 2x)^k+1 > 1 + 2kx + 2x
Therefore, the statement is true for n = k + 1. Thus, the statement is true for all n > 1, n

Originally Posted by GrigOrig99

Can anyone help me confirm if I have solved this correctly?

Many thanks.

Q. (1 + 2x)ⁿ > 1 + 2nx, for x > 0 & n

Attempt: Step 1: For n = 1...
(1 + 2x)¹ = 1 + 2x '&' 1 + 2(1)x = 1 + 2x
Since 1 + 2x > 1 + 2x, n is true.

Step 2: Assume the statement is true for n = k, i.e. assume (1 + 2x)^k > 1 + 2kx,
We must now show that the statement is true for n = k + 1,
i.e. (1 + 2x)^k+1 > 1 + 2(k + 1)x = 1 + 2kx + 2x
(1 + 2x)^k+1 = (1 + 2x)¹(1 + 2x)^k > (1 + 2x)(1 + 2kx)...((1 + 2x)^k > 1 + 2kx...assumed)
If (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x then (1 + 2x)^k+1 > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) > 1 + 2kx + 2x
if (1 + 2x)(1 + 2kx) - (1 + 2kx + 2x) > 0
if 2kx² + 2kx + 2x + 1 - 1 - 2kx - 2x > 0
if 2kx² > 0
if x²[2k] > 0...true when x > 0 & n > 1
(1 + 2x)^k+1 > 1 + 2kx + 2x
Therefore, the statement is true for n = k + 1. Thus, the statement is true for all n > 1, n

I'd do it like this...

Thank you.