mean value theorem?

Method 1

Suppose f(x) < 0 for all x > -1
x - ln(1 + x) < 0
x < ln(1 + x)
ln(1 + x) > x

If x > 0
ln(1 + x) / x > 1
[ln(1 + x) - ln 1] / (x - 0) > 1
1/(1 + x) > 1 ...... ( by mean value theorem )
But 1/(1 + x) < 1 when x > 0
∴ Contradiction

If -1 < x < 0
ln(1 + x) / x < 1
[ln(1 + x) - ln 1] / (x - 0) < 1
1/(1 + x) < 1 ...... ( by mean value theorem )
But 1/(1 + x) > 1 when -1 < x < 0
∴ Contradiction

If x = 0
f(0) = 0 - ln(1 + 0) = 0
∴ Contradiction

∴ f(x) ≥ 0 for all x > -1

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Method 2

f(x) = x - ln(1 + x)
f'(x) = 1 - 1/(1 + x)
f"(x) = 1/(1 + x)²

When f'(x) = 0
1 - 1/(1 + x) = 0
x + 1 = 1
x = 0
f"(0) = 1/(1 + 0)² = 1
∴ f(0) = 0 is the global maximum value

∴ f(x) ≥ 0 for all x > -1

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Method 3

f(x)
= x - ln(1 + x)
= x [x - ln(1 + x) - 0 + ln(1 + 0)] / (x - 0)
= x [1 - 1/(1 + y)] ...... where -1 < y < x ...... ( by mean value theorem )

When -1 < y < x < 0,
1 - 1/(1 + y) < 0 and x < 0
f(x) = x [1 - 1/(1 + y)] > 0

When 0 < y < x
1 - 1/(1 + y) > 0 and x > 0
f(y) = x [1 - 1/(1 + y)] > 0

When x = 0, f(x) = 0

∴ f(x) ≥ 0 for all x > -1