Sol. We have to prove that

(a + b) n ≤ a n + b n for all a, b > 0 and 0 < n < 1.

i.e., ≤ + 1 for all a, b > 0 and 0 < n < 1

So, let us consider the function f(x) given by

f(x) = (1 + x) n –x n –1 for x > 0

Differentiating w.r.t. x, we get

f ′ (x) = n (1 + x) n–1 – nx n–1

f ′ (x) = n

⇒ f ′ (x) < 0

⇒ f(x) is decreasing for all x > 0.

⇒ f(x) ≤ f(0) for all x ≥ 0

⇒ (1 + x) n –x n –1 < 0 for all x ≥ 0

⇒ (1 + x) n < 1 + x n for all x ≥ 0

⇒ < 1 + [Replacing x by a/b]

⇒ (a + b) n < a n + b n Hence, (a + b)

n ≤ a n + b n for all a, b > 0 and

0 < n < 1.