Question

If the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is _______________.

Answer 1

Let the two numbers be x and 4 − x (x ≠ 0, 4).

Suppose S be the sum of their reciprocals.

$\therefore S=\frac{1}{x}+\frac{1}{4-x},x\ne 0,4$

Differentiating both sides with respect to x, we get

$\frac{dS}{dx}=-\frac{1}{x^2}-\frac{1}{{\left(4-x\right)}^2}\times \left(-1\right)$

$\Rightarrow \frac{dS}{dx}=-\frac{1}{x^2}+\frac{1}{{\left(4-x\right)}^2}$

For maxima or minima,

$\frac{dS}{dx}=0$

$\Rightarrow -\frac{1}{x^2}+\frac{1}{{\left(4-x\right)}^2}=0$

$\Rightarrow -\left(16-8x+x^2\right)+x^2=0$

$\Rightarrow 8x=16$

$\Rightarrow x=2$

Now,

$\frac{d^{2}S}{d{x}^2}=\frac{2}{x^3}+\frac{2}{{\left(4-x\right)}^2}$

At x = 2, we have

${\left(\frac{d^{2}S}{d{x}^2}\right)}_{x=2}=$ $\frac{2}{{\left(2\right)}^3}+\frac{2}{{\left(4-2\right)}^2}=$ $\frac{1}{4}+\frac{1}{2}=$ $\frac{3}{4}>0$

So, x = 2 is the point of local minimum.

Thus, S is minimum when x = 2.

∴ Minimum value of S = $\frac{1}{2}+\frac{1}{4-2}$ = $\frac{1}{2}+\frac{1}{2}$ = 1 $\left(S=\frac{1}{x}+\frac{1}{4-x}\right)$

Thus, if the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is 1.


If the sum of two non-zero numbers is 4, then the minimum value of the sum of their reciprocals is ___1___.