# Arithmetic Mean, Geometric Mean, Harmonic Mean and their relations

If $a$ be the arithmetic mean between $b$ and $c$, $b$ be the geometric mean between $c$ and $a$ then prove that $c$ is the harmonic mean between $a$ and $b$. I expressed $a$ as $$a=\frac{(b+c)}{2}$$ $$b=\sqrt {ac}$$ . I solved the equations but I could not evaluate for $c$

• What do you mean you solved if you cannot evaluate for $c$? Commented Aug 20, 2014 at 19:38
• @AlexanderVigodner he tried to solve the equation(he tried to solve for c), but he could not.
– user845875
Commented Aug 2, 2021 at 17:55

We know that $c=2a-b=b^2/a$ ($a$ and $c$ must have the same sign). We have to prove that $$c=\frac{2}{(1/a+1/b)}=\frac{2ab}{a+b}$$ So let's take the difference: $$2a-b-\frac{2ab}{a+b}=\frac{2a^2+2ab-ab-b^2-2ab}{a+b}=\frac{a(2a-b)-b^2}{a+b}$$ Since $b^2=a(2a-b)$ this difference is zero.

Hint: It is a rather well-known and beautiful fact, that the geometric mean of two numbers is also the geometric mean between the arithmetic and harmonic mean of those two numbers:

$\qquad\qquad\quad G=\sqrt{uv}\iff G=\sqrt{AH},~$ where $~A=\dfrac{u+v}2~$ and $~H=\dfrac{2uv}{u+v}$

a is the arithmetic mean between b and c

       a = (b+c)/2 --------------(1)


‘b’ is the geometric mean between a and c

b = √ac  OR b^2= ac - - - - - - - - (2)


by Multiplying eqn(1) by ‘b’ and replacing b2 by eqn (2) in (1) we get,

ab = b ((b+c)/2)

ab = (b2 +bc )/2

ab = (ac +bc )/2

ab = c(a+b) )/2

c = 2ab /(a+b)

hence c is the haromonic mean of a and b

$$a=\frac{b+c}2\tag{AM}$$ $$b=\sqrt{ac}\tag{GM}$$

$$2a=b+c\tag{from AM,(1)}$$ $$2a=\sqrt{ac}+a\tag{from GM b=\sqrt{ac}}$$ $$a=\sqrt{ac}$$ $$\sqrt a=\sqrt c\implies a=c$$ Put $a=c$ in (1) $$2a=b+a\implies b=a$$ So $a=b=c$, so c is the harmonic mean between a and b as: $$\frac2c=\frac1a+\frac1b\tag{since a=b=c}$$

• Error in calculation step 3.
– MonK
Commented Aug 20, 2014 at 19:29
• I don't understand why $b+c=\pm\sqrt{ac}+a$. Where does the $\pm$ come from? And how come $c$ turns into $a$? And at the end, how can you drop $\sqrt{a}$ only on one side? Commented Aug 20, 2014 at 19:47
• @bartgol firstly:$b^2=ac\implies b=\pm\sqrt{ac}$ Commented Aug 20, 2014 at 19:50
• @Aditya $b=\sqrt{ac}$ by the definition. There is no $+-$. Commented Aug 20, 2014 at 20:04
• @bartgol I have proved $a=b=c$ Commented Aug 21, 2014 at 5:15

I think $$a=b=c$$ (and therefore all the means are the same).

Given the two equations, you get

$$ac = b^2 = (2a-c)^2 = 4a^2 - 4ac +c^2$$

Hence

$$c^2 -5ac +4a^2 = 0$$ which, solving for $$c$$, yields $$c=a$$ or $$c=4a$$. Now, since $$b = 2a-c$$, we can only have $$b=a$$ or $$b=-2a$$. Since $$b$$ cannot be negative (otherwise your second assumption would not hold), we get $$b=a$$ too.

• Finally one has to arrive at $$c=\frac {2ab}{a+b}$$ Commented Aug 20, 2014 at 19:47
• If $a=b=c$, then $c=2ab/(a+b)$... Commented Aug 20, 2014 at 19:48
• @bartgol example $a=-1,c=-4,b=2$. Your proof is wrong. Commented Aug 20, 2014 at 20:01
• @AlexanderVigodner you don't allow negative numbers when you deal with geometric means. Otherwise, you either get an imaginary mean (if one is positive) or a mean which is outside the interval. Either way, not fitting our idea of mean. Commented Aug 20, 2014 at 21:10
• I agree generally, but your proof is wrong anyway. Check condition $b=-3a$. Commented Aug 20, 2014 at 21:50

Define $$d_1 := 2a-b-c,\quad d_2 := b^2-ac,\quad d_3 := (a+b)c-2ab$$ where $$\,a,b,c\,$$ are positive real numbers. Now

• $$\,d_1=0\,$$ is equivalent to $$\,a\,$$ being the arithmetic mean of $$\,b\,$$ and $$\,c.$$
• $$\,d_2=0\,$$ is equivalent to $$\,b\,$$ being the geometric mean of $$\,c\,$$ and $$\,a.$$
• $$\,d_3=0\,$$ is equivalent to $$\,c\,$$ being the harmonic mean of $$\,a\,$$ and $$\,b.$$

Verify the simple algebraic identity

$$bd_1 + d_2 + d_3 = b(2a-b-c) + (b^2-ac) + ((a+b)c-2ab) = 0.$$

Given both $$\,d_1 = 0\,$$ and $$\,d_2 = 0,\,$$ then the identity implies that $$\,d_3 = 0.\,$$ This proves the requested result

If $$a$$ be the arithmetic mean between $$b$$ and $$c$$, $$b$$ be the geometric mean between $$c$$ and $$a$$ then prove that $$c$$ is the harmonic mean between $$a$$ and $$b$$.