Suppose that x GE 1 Prove that 1 x2 GE 1 2x Prove that 1

Suppose that x GE -1. Prove that (1 + x)^2 GE 1 + 2x. Prove that (1 + x)^3 GE 1 + 3x. Guess a general inequality for (1 +x)^n when x GE - 1 and n N.

(i)

we are given that x>= -1

and (1 + x)^2 will always be >= 0

and (1 + 2x) >= -1

now we can see that at x = -1

(1 + x)^2 = 0 and (1 + 2x) = -1
and we know that 0 > -1

now at x = 0

(1 + x)^2 = 1 and (1 + 2x) = 1
and we know that 1 = 1

and for all other values of x > 0

(1+x)^2 > (1 + 2x)

hence proved that
for x>=-1

(1+x)^2 >= (1 + 2x)

(ii) for x >= -1

when x = -1

(1 + x)^3 = 0 and (1+3x) = -2

and we know that 0 > -1

when x = 0

(1+x)^3 = 1 and (1+3x) = 1

we know that 1 = 1

and for all other values of x > 0

(1+x)^3 > (1+3x)

hence proved that for x>=-1

(1+x)^3 >= (1+3x)

now if we look at the above two inequalities we could see that

the exponent on the left hand side is the same as the coefficient of x on the right hand side

for exmple in (1+x)^3 >= (1+3x)

the exponent on the left is 3 and the coefficient of x on the right is 3

so in general we could write the inequality as :

(1 + x)^n >= (1 + nx) when x >=-1 and n E N

N stands for natural numbers N = 1 , 2, 3 ,4.......N