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

[anemone]

Find all real solutions to the system of equations below:

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

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

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

 

[Opalg]

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]

 

[anemone]

[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!

 

[juantheron]

My Solution:

Spoiler

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

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

bcz of square quantity in Numerator and Denomenator.

Now Multiply all Three equations, We get

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

Now Using \(\displaystyle \bf{A.M\geq G.M}\).

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

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

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

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

 

[anemone]

My Solution:

Spoiler

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

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

bcz of square quantity in Numerator and Denomenator.

Now Multiply all Three equations, We get

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

Now Using \(\displaystyle \bf{A.M\geq G.M}\).

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

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

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

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

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!