Find a positive number such that the sum of the number and its reciprocal is as small as possible.

Solution:

Let us consider a positive number ‘x’ where x ≠ 0

⇒ Reciprocal = 1/x

Sum of the number and its reciprocal, S = x + 1/x --- (1)

To find minimum value, differentiate (1) w.r.t x and make dS/dx = 0.

dS/dx = 1 - 1/x² --- (2)

Differentiating (2) w.r.t x,

d²S/dx² = 0 - [(-2)x ^-3] = 2/x³

Now, for S to have minimum value dS/dx = 0

⇒1 - 1/x² = 0

1/x² = 1

x² = 1

x = ±1

x = -1 is not possible as x ＞0

Therefore, the value of x =1

Now when x =1, d²S/dx² = 2/x³ = 2/1² = 2 ＞0

S is minimum when x = 1

Therefore, the minimum value is value of S when x = 1 {Equation(1)}

S = x + 1/x

S = 1 + 1/1 

S = 2

Therefore, the required numbers are 1 and 1

Aliter

Let S = x + 1/x

We have, the relation between arithmetic mean (AM) and the geometric mean (GM) of two positive real numbers ‘a’ and ‘b’ as,

AM ≥ GM

(a + b)/2 ≥ √(a × b)

⇒ [x + (1/x)]/2 ≥ √(x × (1/x))

[x + (1/x)]/2 ≥ 1

x + (1/x) ≥ 2

⇒ S ≥ 2

The minimum value of S is 2.

Therefore, the required numbers are 1 and 1

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Find a positive number such that the sum of the number and its reciprocal is as small as possible.

Summary:

Positive numbers such that the sum of the number and its reciprocal, which is as small as possible, are 1 and 1.