Zeros at the same time

We assume that both are zero. Then we have :

$\frac {1} {a} + \frac {1} {b} + \frac {1} {c} = 0 \iff \frac {ab+ac+cb}{abc} = 0 \iff ab+ac+cb=0$

We also have : $a+b+c=0 \iff (a+b+c)^{2}=0 \iff a^2+b^2+c^2+2(ab+ac+bc)=0 \iff a^2+b^2+c^2=0$ Which is a contradiction ! Hence those numbers cannot be both zero at the same time

Similar with this

If $a+b+c=0$ ,then $c=-(a+b),\frac1a+\frac1b+\frac1c=\frac1a+\frac1b+\frac1{-(a+b)}=0,\frac1a+\frac1b=\frac1{a+b}$ which is impossible.

We have $\dfrac1a+\dfrac1b+\dfrac1c=\dfrac{ab+bc+ca}{abc}=0$ which says $ab+bc+ca=0$ .

By assuming that $a,b,c$ are real roots of a cubic equation, then we have $\begin{cases}a+b+c=0\\ab+bc+ca=0\\abc=k\end{cases}$ for some real value of $k$ .

These three roots satisfying equation $x^3-k=0$ , which has roots $\sqrt[3]{k}, \omega\sqrt[3]{k}, \omega^2\sqrt[3]{k}$ where $\omega$ is third root of unity. If $k$ is non-zero, it will contradict to the fact that roots are all real, but if $k$ is zero, then all of the roots are zeros, which is also a contradiction. Hence, the answer is $\boxed{\text{No}}$ .

We have a+b+c = 0

Assume a and b are positive and c is negative.

Thus a+b = -c (other assumptions could be made, but they are all symmetrically equivalent)

thus $\frac{1}{a}$ + $\frac{1}{b}$ + $\frac{1}{c}$ = 0

gives $\frac{1}{a}$ + $\frac{1}{b}$ = - $\frac{1}{c}$

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

$\frac{ab}{a+b}$ = -c

combining: a+b = $\frac{ab}{a+b}$

$(a+b)^2 = ab$ Thus ab is positive

$a^2+2ab+b^2 = ab$

$a^2+b^2 = -ab$ This is a contradiction.

First off, let's take off one variable: $a=-b-c$

With that out of the way, we have $\frac 1 {-b-c} + \frac 1 b + \frac 1 c=0$ which can be manipulated to $b^2+bc+c^2=0$

Applying the quadratic formula trying for $b$ we get $\frac{-c\pm\sqrt{c^2-4c^2}} 2$

But to solve that quadratic formula either $c$ must be imaginary or we get an imaginary result for $b$ .

Therefore the answer is no, it's not possible.

Assume there exist such $a, b, c$ . Then we have $c=-(a+b)$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a+b}{ab}-\frac{1}{a+b}=0$ WLOG, since obviously $a, b, c$ can't all negative and positive, we can pick $a$ and $b$ such that they share the same sign

Simplify, we have $(a+b)^2=ab$ $|a+b|=\sqrt{ab}$ However, by AM-GM Inequality and the fact that $a$ and $b$ have same sign, we have $|a+b|=|a|+|b|\ge 2\sqrt{|a||b|}>\sqrt{ab}$ Therefore the assumption leads to a contradiction, so there do not exist such values $a, b, c$

My solution involves logic. We start by observing one or two of the integers must be negative. We can focus on the case where only is integer is negative because of symmetry. Let's say that c is negative and a and b positive. We then have (with d=-c):

$a+b=d$

$\frac{1}{a}+ \frac{1}{b} = \frac{1}{d}$

Because everything is positive, we can derive from the first equation that d is greater than a and b, and from the second equation, we get that d is smaller than a and b. The only possibility is that a=b=d, but that doesn't work out. So it's impossible.

Suppose both are zero. Then $c=-(a+b)$ and $\dfrac{1}{c}=-\dfrac{1}{a}-\dfrac{1}{b}=-\dfrac{a+b}{ab}=\dfrac{c}{ab}$ . This means that $c^2=ab$ . Now considering that $c=-(a+b)$ , we have $(a+b)^2=ab$ . Considering that $(a+b)^2$ is positive, $ab$ must also be positive. But on the other hand, upon subtracting $4ab$ form both sides, $(a-b)^2=-3ab$ . Considering that $ab$ is positive, $(a-b)^2$ would then be negative!! This is impossible!!

We assume that:

a + b + c = 0 = a^(-1) + b^(-1) + c^(-1)

Adding both equations we recieve:

a + a^(-1) + b + b^(-1) + c + c^(-1) = 0

Since any real number squared is always non-negative we can assume:

(a - 1)^2 >= 0

a^2 - 2a + 1 >= 0

a - 2 + a^(-1) >= 0

(We allowed to divide by a because a/=0)

a + a^(-1) >= 2

The same is also true for b and c.

Hence a + a^(-1) + b + b^(-1) + c + c^(-1) >= 6 and our assumption must be false.

(I am quite new here and don't know how to include fractions etc., so sorry for that)

Write $c = nb$
The equations then say:
$a+b(n+1) = 0$ and
$ab+anb+nb^2 = 0 <=> a(n+1) + b = 0$
=> $n+1 = -\frac{b}{a} = -\frac{a}{b}$
=> $-(n+1)ab = a = b$ . So b can be ommited, leaving the equations as
$2a +na = a (2+n) = 0$ and
$2 (\frac{1}{a} + \frac{1}{na}) = \frac{1}{a} (2+\frac{1}{n})= 0$ $<=>$ (with $a \neq 0$ ):
$2+n = 0$ and $2+\frac{1}{n} = 0$

Equation 1:

$a+b=-c$

---

Equation 2:

$\frac{1}{a} + \frac{1}{b} = \frac{-1}{c}$

$-c = \frac{1}{1/a+1/b}$

---

Equalizing equations:

$a+b = \frac{1}{1/a+1/b}$

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

$\frac{1}{a+b} = \frac{a+b}{a.b}$

$ab = (a+b)^2$

$a^2+b^2+2ba = ab$

$a^2+ba+b^2 = 0$

$c = \frac{-b \pm \sqrt{b^2-4b^2}}{2}$

$\sqrt{b^2-4b^2} = \sqrt{-3b^2}$

$\nexists b \in R , b^2<0$

$a \notin R$

Contradiction

1) a+b=-c, 2)1/a + 1/b = -1/c ==> ab = cc Square 1) in both sides: (a+b)(a+b) = aa+2ab+bb = cc ==> aa+ab+bb=0 (a+1/2b)(a+1/2b) + 3/4bb = 0

So, b = 0 ==> a = 0 ==> c = 0 Thus, contradiction.

Without loss of generality, we can assume that $a$ and $b$ are positive and $c$ is negative. Then $a+b=-c=|c|$ .

In particular, $0 < a, b < |c|$ . But then $\frac1a,\frac1b>\frac 1{|c|}$ .

Therefore $\frac1a+\frac1b+\frac 1c = \frac1a+\frac1b-\frac 1{|c|} > \frac1a + 0>0$ , and hence $\frac1a+\frac1b+\frac 1c\ne0$

If a+b+c = 0

a+b+c>1/(1/a+1/b+1/c) [ am hm inequality]

Hence 1/(1/a +1/b+1/c) is not equal to 0

If there is a real solution, 1/a + 1/b = 1/(a + b). Multiplying through by (a + b): -> 1+ b/a + a/b + 1 = 1 Simplifying and multiplying through by a: -> a^2/b + a + b = 0 Solving for a: -> a = { -1 +/- sqrt (1 - 4b/b) } / (2/b)

Therefore a has no real solution since it involves the square root of -3.

Consider the polynomial $x^3 + px^2 + qx + r = 0$ and let a,b,c be its roots .Then, the conditions suggest that p = 0 and q = 0 .so the polynomial becomes $x^3 = -r$ . Since $f(x) = x^3$ is an increasing function there exist only 1 real root of it.

$a=-(b+c)$ and $\frac{1}{b}+\frac{1}{c}=\frac{1}{b+c}$ therefore:

$\frac{b+c}{bc}=\frac{1}{b+c}$ wich is: $(b+c)^2=bc$

$(b+c)^2\geq0$ therefore $bc\geq0$

If $(b+c)^2=bc=0$ then $b=0$ or $c=0$ wich is false, so in other words $(b+c)^2=bc>0$

Manipulating this $(b+c)^2=bc$ expression you get $b^2+c(b+c)=0$ , $b^2$ is always positive so $c(b+c)$ has to be negative.

That leads to:

$b+c<0$ for this to be true you have to have $b<-c$ . This statement contradict with fact that $b>0$ because $-c$ is always negative.

So there is no solution for the problem!!!

Not the simplest solution, but I like it:

Let a,b,c be solutions to a cubic equation. Since a+b+c=0, the coefficients of the x^2 term in the cubic is zero.

The second property means ab+ac+bc=0, so the coefficient of the x term in the cubic is zero.

Hence, a,b,c are solutions to a cubic of the form mx^3+n=0, where m and m are constants.

Since this cubic only has one real solution (obvious if you consider the graph) then a,b,c must all be the same, so the only way a+b+c=0 is if they are all zero.

Bcuz 1/a cannot equal to 0

We can see that an opposite sign number in the equation is necessary to cancel the values of the other two numbers. For example in $a + b + c = 0$ we can have:

$1 + 2 - 3 = 0$

$-1 + (-2) + 3 = 0$

Now we know that not only we need a number with an opposite sign but also that this number must be the greatest in the case of $a + b + c = 0$ .

The equation $\frac {1}{a} + \frac{1}{b} + \frac{1}{c} = 0$ , can be solved independently with several values:

$\frac {1}{3} + \frac{1}{6} + \frac{1}{-2} = 0$

$\frac {1}{-12} + \frac{1}{-4} + \frac{1}{3} = 0$

But this time is the number (a, b, c) with the least value that needs to have the opposite sign, which creates a contradiction with the first equation.

Therefore there is no solution for any real number different than zero.

We are given $c = -(a + b)$ and $\frac{1}{a} + \frac{1}{b} = - \frac{1}{c}$

So we have $\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}$

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

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

$\Rightarrow a^2+2ab+b^2 = ab$

$\Rightarrow a^2+ab+b^2 = 0$

Now if we try to find the roots of the above equation for either $a$ or $b$ , we can see there are no real roots.

That is,

$a = \frac{-b\pm\sqrt{b^2-4b^2}}{2} = \frac{-b\pm\sqrt{-3b^2}}{2}$

So, $a$ cannot be real.

I did it differently. It is abvious a,b,and c cannot have the same sign as if the did a+b+c would be non zero.

Let a and b share the same sign. Then c has the opposite sign and c= -(a+b)

So the sum of the inverses is 1/a + 1/b - 1/(a+b) = 0

Multiply through by ab(a+b) So b(a+b)+ a(a+b)- ab=0 So a^2+ b^2+ ab=0 and since all three of those items are non zero and positive(since a and b have the same sign), we have a contradiction.

So it's impossible.

Assume there exists non-zero real numbers $a$ , $b,$ $c$ such that $a+b+c = 0$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$

$\implies$

$b + c = -a$ and $b + c = -\frac{bc}{a}$

$b + c = -a \implies c = -a - b \implies b^2 + ab + a^2 = 0 \implies b = a(-\frac{1}{2} \pm \frac{\sqrt{3}}{2}i)$

$a$ is a nonzero real number $\implies b \in \mathbb{C}$ and $c \in \mathbb{C}$ .

$\therefore$ there does not exist non-zero real numbers $a$ , $b,$ $c$ such that $a+b+c = 0$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$

By the given conditions it is found that $a^{2}+b^{2}+c^{2}=0$ and also in the question it says that a, b and c are non-zero. But the sum of squares cant be zero. So values for a,b,c do not $EXIST$

e1:

$a + b = -c$

e2:

$-c = \frac{1}{\frac{1}{a} + \frac{1} {b}}$

e1 = e2:

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

To pq form:

$b^2 + ab + a^2 = 0$

The term under the root is <0 so a, b or c must be 0.

I know thats not the question but it should work if:

$b = -\frac{a} {2} \pm \frac{\sqrt{3}}{2}ai$

And

$c = -a - b$

So if two of three numbers are irrational?

Similar to Kelvin Hong, we can deduce that a,b,c are roots to some cubic equation of the form x^3 = k. We can see that because the equation y = x^3 has a well-defined inverse, this equation will have one solution for all real numbers k. Thus a,b,c cannot be all real.

Suppose: (1) a + b + c = 0; (2) 1/a + 1/b + 1/c = 0.Multiply ( 2 ) by abc, resulting in (3) bc + ac + ab + 0; multiply (1) by a, resulting in (4) a^2 + ab + ab = 0. Equating (3) and (4), (5) a^2 = bc. But a^2 = b^2 + 2bc + c^2, and substituting, a^2 + b^2 + c^2 = 0; Then a = b = c = 0, and the right side is undefined. Ed Gray

Assume $a+b+c=0$ and $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$ .

The second equation is only defined for $b\neq 0$ and both equations are homogeneous, so we may set $b=1$ . Then by the first equation, $c = -1 - a$ . Putting this into the second equation gives $\frac{1}{a} + 1 = \frac{1}{1+a} \implies a^2 + a + 1 = 0 \implies a = \frac{-1 \pm i\sqrt{3}}{2}$ which is not real.