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The question is:

Determine the maximum value of $a, b, c$ from the following system of quadratic equations. $$\frac{4a^2}{1+4a^2}=b$$ $$\frac{4b^2}{1+4b^2}=c$$ $$\frac{4c^2}{1+4c^2}=a$$

I haven't an idea on how to do this, and my attempts led to a dead end. Help would be appreciated, thanks.

ryan.zcd
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  • You could start by first pluggin in the third equation into the first one, i.e. instead of writing $a$, write $\frac{4c^2}{1+4c^2}$, and use the second equation to subsitute $c$ by the expression for $b$. This leaves you with just an equation for $b$, which then can be used in order to calculate $a$, $b$. – Nuke_Gunray Mar 10 '23 at 01:56
  • @Nuke_Gunray that will lead to something like $b {{\left( 2 b-1\right) }^{2}} \left( 5696 {{b}^{6}}+1600 {{b}^{5}}+496 {{b}^{4}}+96{{b}^{3}}+28{{b}^{2}}+4 b+1\right)=0$ with the obvious real roots of $0$ and $\frac12$ (as a double root). The final term is clearly positive for non-negative $b$ (in fact for all $b$ but that is not needed) so the maximum real root is $\frac12$. Similarly for $a$ and $c$, and $a=b=c=\frac12$ is a solution and so the maximal real one – Henry Mar 10 '23 at 11:01

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$\frac{4a^2}{1+4a^2}=b \wedge \frac{4b^2}{1+4b^2}=c \Rightarrow c = \frac{64 a^4}{80 a^4 + 8 a^2 + 1}$, while $\frac{4c^2}{1+4c^2}=a$.
Now, we substitute the last one into the second term of $c = \frac{64 a^4}{80 a^4 + 8 a^2 + 1}$, so we finally get $c(2c-1)=0 \Rightarrow c \in \{0,\frac{1}{2}\}$ and hence, $max\{0,\frac{1}{2}\}=0.5$.
Thus, by symmetrical construction, you can easily show that $max\{a\}=max\{b\}=max\{c\}=max\{0,\frac{1}{2}\}=0.5$ (and similarly $min\{a\}=min\{b\}=min\{c\}=min\{0,\frac{1}{2}\}=0$).

Marco Ripà
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    How do you get $c(2c-1)=0$ rather than $c {{\left( 2 c-1\right) }^{2}} \left( 5696 {{c}^{6}}+1600 {{c}^{5}}+496 {{c}^{4}}+96{{c}^{3}}+28{{c}^{2}}+4 c+1\right)=0$? – Henry Mar 10 '23 at 11:03
  • Let $c$ belong to the set $\mathbb{R}$ by hypothesis. By substitution, we get $c = \frac{16384 c^8}{22784 c^8 + 1280 c^6 + 224 c^4 + 16 c^2 + 1}$ which has only two real solutions (i.e., $c=0$ and $c=\frac{1}{2}$), thus $2c^2-c=0$ (Q.E.D.) – Marco Ripà Mar 10 '23 at 14:56
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A clear solution is $a = b = c = 0$. Assuming that is not the case, one can rewrite the equations as: \begin{align*} \frac{1}{b} = 1 + \frac{1}{4a^2}; \quad \frac{1}{c} = 1 + \frac{1}{4b^2}; \quad \frac{1}{a} = 1 + \frac{1}{4c^2}. \end{align*} On adding them all up, we obtain, \begin{align*} 3 + \frac{1}{4a^2} + \frac{1}{4b^2} + \frac{1}{4c^2} & = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \\ \implies \frac{1}{4a^2} -\frac{1}{a} + 1 + \frac{1}{4b^2} -\frac{1}{b} + 1 + \frac{1}{4c^2} -\frac{1}{c} + 1 & = 0\\ \left(\frac{1}{2a} - 1\right)^2 + \left(\frac{1}{2b} - 1\right)^2 + \left(\frac{1}{2c} - 1\right)^2 & = 0. \end{align*} The only solution to the above equation is that all the terms within the brackets are $0$, which yields us $a = b = c = 1/2$.

sudeep5221
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  • My answer below includes it, indeed. Anyway, I think that the search is over, since we have that $a=0 \Rightarrow b=0 \Rightarrow c=0$, whereas $a=0.5 \Rightarrow b=0.5 \Rightarrow c=0.5$. – Marco Ripà Mar 10 '23 at 02:37
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    Thanks for pointing it out @JohnOmielan. Updated the solution. – sudeep5221 Mar 10 '23 at 04:08
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$\def\eqdef{\stackrel{\text{def}}{=}}$ Here's another approach.

Define $\ f:\mathbb{R}_{\ge0}\rightarrow \mathbb{R}_{\ge0}\ $ by $$ f(x)\eqdef\frac{4x^2}{1+4x^2}\ . $$ If $\ x_f\ $ is a fixed point of $\ f\ $ (that is, $\ f(x_f)=x_f\ $) then $\ a=$$\,b=$$\,c=$$\,x_f\ $ is a solution for the three equations. One fixed point of $\ f\ $ is obviously $\ x_f=0\ $, and any non-zero fixed point, $\ x_f\ $, of $\ f\ $ must satisfy $$ \frac{4x_f}{1+4x_f^2}=1\ , $$ or equivalently, $$ (2x_f-1)^2=0\ . $$ Thus $\ x_f=0\ $ and $\ x_f=\frac{1}{2}\ $ are the only fixed points of $\ f\ $.

If $\ a,b,c\ $ satisfy the three equations then they must each be fixed points of $\ f^3\ $. But since $$ f(x)-x=\frac{-x\big(2x-1\big)^2}{1+4x^2}<0 $$ for all non-negative $\ x\ne0,\frac{1}{2}\ $, then $\ f^3(x)<x\ $ also, for all non-negative $\ x\ne0,\frac{1}{2}\ $. Therefore $\ f^3\ $ has no other fixed points than those of $\ f\ $, and $\ a=b=c=0\ $ and $\ a=b=c=\frac{1}{2}\ $ are the only solutions of the three equations.

lonza leggiera
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  • The observation that $f(x) - x \leq 0 $ is crucial, and you might want to highlight that instead of making the focus just on fixed points. – Calvin Lin Mar 10 '23 at 20:11