Denominator Summation

$\dfrac{1}{a+b} = \dfrac{1}{a} + \dfrac{1}{b}\\ \dfrac{1}{a+b} = \dfrac{a+b}{ab}\\ (a+b)^2 = ab\\ a^2+2ab+b^2 - ab = 0\\ a^2 + ab + b^2 = 0$

Now, we let $b$ be a positive integer constant, and $a$ be the variable. This gives us a quadratic equation. The discriminant,

$\Delta =b^2 - 4(1)(b^2) =-3b^2$

Since we know that $b>0$

$b^2 > 0\\ -3b^2 < 0\\ \Delta < 0$

This means that for any positive integer of $b$ , we won't get a real value for $a$ .

Therefore, no positive integers $a$ and $b$ satisfy $\dfrac{1}{a+b} = \dfrac{1}{a} + \dfrac{1}{b}$

The answer is $\boxed{\text{False}}$

For this question both $a$ and $b$ are positive integers. Therefore $a>0,b>0$

From this we can see that $a+b > a,a+b > b$

For any two numbers the inverse of the larger number is always less than the inverse of the smaller number. For example, $8>2$ and $\frac{1}{8} < \frac{1}{2}$

So we can conclude that since $a+b$ is greater than both $a$ and $b$ , then $\frac{1}{(a+b)}$ is less than both $\frac{1}{a}$ and $\frac{1}{b}$

This shows that for any case, $\frac{1}{(a+b)} < \frac{1}{a} + \frac{1}{b}$ showing that $\boxed{\textbf{FALSE}}$ is the answer to the question .

Multiply both sides by a+b

$\frac {a+b}{a+b} = \frac {a+b}{a} + \frac {a+b}{b}$

LHS = 1

Both a and b are > 0 therefore both RHS fractions are > 1.

So LHS = 1, RHS > 2, therefore there is no solution.

Let us assume that the given statement is true.

$\implies \frac{a+b}{ab} = \frac{1}{a+b}$

$\implies (a+b)^2=ab$

$\implies (a+b)=\sqrt{ab}$

$\implies \frac{a+b}{2} < \sqrt{ab}$

But by AM-GM Inequality we have,

$\frac{a+b}{2} \geq \sqrt{ab}$

This contradicts are assumption. Hence are assumption is incorrect.

By solving we get a^2+b^2=-ab Which is never possible as sum of positive numbers cannot be negative

A simple solution would be multiplying by (a+b) giving 1=b+a

In which there's no positive integer pair that can add to one.

Removing fractions

ab=b(a+b) + a(a+b)

a^2 + b^2 + ab = 0

Since a and b are both positive this is impossible therefore the statement is false