Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$

Question

Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $ \sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$

Here is my attempt:

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$

I am having trouble in figuring out the best approach to simplify $ \sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$
so that I can find the value of $ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Hope somebody has an idea.

Answer 1

Solving your equation $$\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$$ for $c$ we get $$c=-\frac{ab}{a+b}$$ then we get

$$\frac{1}{a}+\frac{1}{b}-\frac{a+b}{ab}=0$$

Answer 2

Square both sides to get

$$b=-\sqrt{(a+b)(b+c)}$$

square again

$$0=ab+ac+bc$$

divide by $abc$ and finish.

Answer 3

Take the square of the condition : $$a+b+b+c+2\sqrt{a+b}\sqrt{b+c}=a+c$$ This implies $b=-\sqrt{a+b}\sqrt{b+c}$, so $b^2=(a+b)(b+c)$, hence $$ab+bc+ac=0$$ So your answer should be $0$ (what else could it be ?) :-)

EDIT :

I thought about this a little (well, OK, a big !) more, and especially the feasibility of the initial condition.

We have $(1)\,:\,ab+ac+bc=0$, so $a$, $b$ and $c$ can't be all positive. Which asks the question : can $a+b$, $a+c$ and $b+c$ be positive ?

Now we can't have two negative values, because the sum would be negative. So only one of the three variables can be negative. And it can't be $a$ or $c$, because, for example, if $a<0$, $a+c<c<b+c$, and the left member of the condition is strictly greater than the right member.

So $b<0$, and solving for $b$ in $(1)$, we find $b=-\frac{ac}{a+c}=-\frac{1}{1/a+1/c}$ (which, BTW, is half the opposite of the harmonic mean of $a$ and $c$).

Remains to prove that $a+b$ and $b+c$ are both positive. For example $$a-\frac{ac}{a+c}=\frac{a^2}{a+c}>0$$ Now I can sleep without thinking anymore at this funny problem :-)

END EDIT.

And BTW : welcome to MSE !

Answer 4

Just to do it the hard way.

Let $m = a+b$ and $n = b+c$ and $k = a + c$.

$\sqrt m + \sqrt n = \sqrt k$ so $k = m + n + 2\sqrt mn$ and

$\frac 1a + \frac 1b + \frac 1c = \frac 2{m - n +k} + \frac 2{n-k + m} + \frac 2{k-m + n}$

$=2(\frac {(n-k+m)(k-m + n) + (m-n+k)(k-m+n)+ (m-n+k)(n-k+m)}{(m-n+k)(n-k+m)(k-m+n)})$

$n-k +m = -2\sqrt{mn};$ $k-m+n = 2(n + \sqrt{mn});$ and $m-n+k = 2(m + \sqrt{mn})$

so we have

$\frac 1a + \frac 1b + \frac 1c=\frac {2*4}{8} \frac {-\sqrt{mn}(n+\sqrt{mn}) + (m+\sqrt{mn})(n+\sqrt{mn}) -\sqrt{mn}(m+\sqrt{mn})}{-\sqrt{mn}(n+\sqrt{mn}) (m + \sqrt{mn})}$

$=\frac {-n\sqrt{mn} - mn + mn +n\sqrt{mn} + m\sqrt{mn} + mn - m\sqrt{mn} - mn}{nm\sqrt{nm} + m^2n + mn^2 + mn\sqrt{mn}}$

$= \frac {0}{nm\sqrt{nm} + m^2n + mn^2 + mn\sqrt{mn}} = 0$

====

The "other" hard way to do it:

$\sqrt{a + b} + \sqrt{b + c} = \sqrt{a+c}$. So it seems reasonable we can attempt to solve $b$ in terms of $a $ and $c$

$(a + b) + (b+c) + 2\sqrt{a+b}\sqrt{b+c}= a + c$ so

$\sqrt{a+b}\sqrt{b+c}= - b$ (so $b \le 0$)

So $(a+ b)(b+c) = b^2$ and $ab + b^2 +ac + bc = b^2$ so $b(a+c) = -ac$

If $a + c \ne 0$ then $b = \frac {-ac}{a+c}$.

If $a +c = 0$ then $ac = 0$ and either $a$ or $c$ equal $0$ and we were told they were non-zero.

So $b = \frac {-ac}{a+ c}$.

So $\frac 1a + \frac 1b + \frac 1c = \frac 1a - \frac {a+c}{ac} + \frac 1c= \frac c{ac} - \frac {a+c}{ac}+ \frac a{ac} = \frac {a -(a+c) +c}{ac} $ and .... well, now it just falls out. It's almost ridiculous.

....

But it's interesting to note that $b$ is negative and $a$ and $c$ are positive.

It makes one wonder just how obvious it should be that $\sqrt{a - \frac 1{\frac 1a + \frac 1c}} + \sqrt{c - \frac 1{\frac 1a + \frac 1c}} =\sqrt{a + c}$?