If the sum of two non-zero numbers is 4 ,

Solution:

Let the two numbers be $x$ and $4-x(x \neq 0,4)$.

Suppose S be the sum of their reciprocals.

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

Differentiating both sides with respect to x, we get

$\frac{d S}{d x}=-\frac{1}{x^{2}}-\frac{1}{(4-x)^{2}} \times(-1)$

$\Rightarrow \frac{d S}{d x}=-\frac{1}{x^{2}}+\frac{1}{(4-x)^{2}}$

For maxima or minima,

$\frac{d S}{d x}=0$

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

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

$\Rightarrow 8 x=16$

$\Rightarrow x=2$

Now, 

$\frac{d S}{d x}=0$

At $x=2$, we have

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

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

Thus, S is minimum when x = 2.

$\therefore$ 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___.