MHB Real Solutions to System of Equations: a, b, and c Values

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The system of equations defined by a, b, and c leads to the conclusion that the only real solutions are a = b = c = 0 and a = b = c = 0.5. The function f(x) = 4x^2/(1 + 4x^2) is shown to be bounded within the interval [0, 1) and has no periodic points of order 3. Analysis reveals that f(x) is always less than or equal to x for x in the unit interval. The discussion highlights the use of mathematical tools, such as the AM-GM inequality, in solving these types of equations.
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Find all real solutions to the system of equations below:

$a=\dfrac{4c^2}{1+4c^2}$

$b=\dfrac{4a^2}{1+4a^2}$

$c=\dfrac{4b^2}{1+4b^2}$
 
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anemone said:
Find all real solutions to the system of equations below:

$a=\dfrac{4c^2}{1+4c^2}$

$b=\dfrac{4a^2}{1+4a^2}$

$c=\dfrac{4b^2}{1+4b^2}$
[sp]Let $f(x) = \dfrac{4x^2}{1+4x^2}$. Clearly $0\leqslant f(x) <1$. Therefore $a,b,c$ all lie in the interval $[0,1).$

Next, $x - f(x) = x - \dfrac{4x^2}{1+4x^2} = \dfrac{x + 4x^3 - 4x^2}{1+4x^2} = \dfrac{x(1-2x)^2}{1+4x^2}.$ That is zero when $x=0$ or $x=\frac12$, and it is strictly positive through the remainder of the unit interval. Thus $f(x) \leqslant x$ for all $x$ in the unit interval, and $f$ can have no periodic points of order $3$. In fact, if $f^{(3)}$ denotes the composition $f\circ f\circ f$, then $f^{(3)}(x) \leqslant (f\circ f)(x) \leqslant f(x) \leqslant x.$ So the only way in which we can have $f^{(3)}(x) = x$ is if $f(x) = x$.

In terms of $a,\,b$ and $c$, this says that the only solutions to the given equations are $a=b=c=0$ and $a=b=c=\frac12.$[/sp]
 
Opalg said:
[sp]Let $f(x) = \dfrac{4x^2}{1+4x^2}$. Clearly $0\leqslant f(x) <1$. Therefore $a,b,c$ all lie in the interval $[0,1).$

Next, $x - f(x) = x - \dfrac{4x^2}{1+4x^2} = \dfrac{x + 4x^3 - 4x^2}{1+4x^2} = \dfrac{x(1-2x)^2}{1+4x^2}.$ That is zero when $x=0$ or $x=\frac12$, and it is strictly positive through the remainder of the unit interval. Thus $f(x) \leqslant x$ for all $x$ in the unit interval, and $f$ can have no periodic points of order $3$. In fact, if $f^{(3)}$ denotes the composition $f\circ f\circ f$, then $f^{(3)}(x) \leqslant (f\circ f)(x) \leqslant f(x) \leqslant x.$ So the only way in which we can have $f^{(3)}(x) = x$ is if $f(x) = x$.

In terms of $a,\,b$ and $c$, this says that the only solutions to the given equations are $a=b=c=0$ and $a=b=c=\frac12.$[/sp]

Brilliant!(Yes)(Yes) Thanks for your intelligent solution and thanks for participating, Opalg!
 
My Solution:

Given $$\displaystyle a = \frac{4c^2}{1+4c^2}$$ and $$\displaystyle b= \frac{4a^2}{1+4a^2}$$ and $$\displaystyle c=\frac{4b^2}{1+4b^2}$$

Clearly here $$\left(a,b,c\right)=(0,0,0)$$ now if $$\left(a,b,c\right)\neq (0,0,0)$$ then here $$a,b,c>0$$

bcz of square quantity in Numerator and Denomenator.

Now Multiply all Three equations, We get

$$\Rightarrow \displaystyle \frac{64(abc)^2}{\left(1+4a^2\right)\left(1+4b^2\right)\left(1+4c^2\right)} = abc\Rightarrow \left(1+4a^2\right)\cdot \left(1+4b^2\right)\cdot \left(1+4c^2\right) = 64abc$$

Now Using $$\bf{A.M\geq G.M}$$.

$$\left(1+4a^2\right)\geq 4a$$ and $$\left(1+4b^2\right)\geq 4b$$ and $$\left(1+4c^2\right)\geq 4c$$

Now Multiply all three, We get $$\left(1+4a^2\right)\cdot \left(1+4b^2\right)\cdot \left(1+4c^2\right)\geq 64abc$$

and equality hold when $$\displaystyle a = b=c = \frac{1}{2}$$

So the solution of the equation are $$\displaystyle \left(q,b,c\right) = \left(0,0,0\right)$$ and $$\displaystyle \left(q,b,c\right) = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$$
 
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jacks said:
My Solution:

Given $$\displaystyle a = \frac{4c^2}{1+4c^2}$$ and $$\displaystyle b= \frac{4a^2}{1+4a^2}$$ and $$\displaystyle c=\frac{4b^2}{1+4b^2}$$

Clearly here $$\left(a,b,c\right)=(0,0,0)$$ now if $$\left(a,b,c\right)\neq (0,0,0)$$ then here $$a,b,c>0$$

bcz of square quantity in Numerator and Denomenator.

Now Multiply all Three equations, We get

$$\Rightarrow \displaystyle \frac{64(abc)^2}{\left(1+4a^2\right)\left(1+4b^2\right)\left(1+4c^2\right)} = abc\Rightarrow \left(1+4a^2\right)\cdot \left(1+4b^2\right)\cdot \left(1+4c^2\right) = 64abc$$

Now Using $$\bf{A.M\geq G.M}$$.

$$\left(1+4a^2\right)\geq 4a$$ and $$\left(1+4b^2\right)\geq 4b$$ and $$\left(1+4c^2\right)\geq 4c$$

Now Multiply all three, We get $$\left(1+4a^2\right)\cdot \left(1+4b^2\right)\cdot \left(1+4c^2\right)\geq 64abc$$

and equality hold when $$\displaystyle a = b=c = \frac{1}{2}$$

So the solution of the equation are $$\displaystyle \left(q,b,c\right) = \left(0,0,0\right)$$ and $$\displaystyle \left(q,b,c\right) = \left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$$

Good job jacks!(Yes) I like it when the AM-GM inequality could be used as a tool to solve for any appropriate system of equations, and thanks for participating!
 
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