Non-linear System of Equations with 3 Variables

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Non-linear System of Equations with 3 Variables

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Source: European Mathematical Cup 2012, Junior Division, Problem 3

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#1 h Jul 27, 2013, 12:59 AM • 2 Y

Y by Adventure10, Mango247

Are there positive real numbers $x$ , $y$ and $z$ such that

$x^4 + y^4 + z^4 = 13\text{,}$

$x^3y^3z + y^3z^3x + z^3x^3y = 6\sqrt{3} \text{,}$

$x^3yz + y^3zx + z^3xy = 5\sqrt{3} \text{?}$

Proposed by Matko Ljulj.

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#2 h Jul 27, 2013, 9:54 AM • 1 Y

Y by Adventure10

Matematika wrote:

Are there positive real numbers $x$ , $y$ and $z$ such that
$x^4 + y^4 + z^4 = 13\text{,}$

$x^3y^3z + y^3z^3x + z^3x^3y = 6\sqrt{3} \text{,}$

$x^3yz + y^3zx + z^3xy = 5\sqrt{3} \text{?}$

My solution

Given is equivalent to

$\begin{cases}x^4 + y^4 + z^4 = 13\\ xyz(z^2y^2+y^2z^2+z^2x^2)= 6\sqrt{3}\\ xyz(x^2+y^2+z^2)= 5\sqrt{3}\end{cases}$

From second equation and third equation we get

$\dfrac{xyz(x^2y^2+y^2z^2+z^2x^2)}{xyz(x^2+y^2+z^2)}=\frac{6\sqrt{3}}{5\sqrt{3}}$

Or $5(x^2y^2+y^2z^2+z^2x^2)=6(x^2+y^2+z^2)$

One other hand, note that $(x^2+y^2+z^2)^2=(x^4+y^4+z^4)+2(x^2y^2+y^2z^2+z^2x^2)$

So we get
$(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+z^2x^2)=13$

Now we have a new system of equations

$\begin{cases}u^2-2v=13\\ 5v=6u\end{cases}$

(Where $u=x^2+y^2+z^2$ and $v=x^2y^2+y^2z^2+z^2x^2$ )

So we get $x^2+y^2+z^2=5$ . So we have $x^2y^2+y^2z^2+z^2x^2=6$ and $xyz=\sqrt{3} \implies x^2y^2z^2=3$ .
Or
$\begin{cases}x^2+y^2+z^2=5\\x^2y^2+y^2z^2+z^2x^2=6\\x^2y^2z^2=3 \end{cases}$

Now we easy to solve and i found

$x=y=z=\sqrt{\frac{1}{3} (5+\sqrt[3]{\frac{61}{2}-\frac{9\sqrt{29}}{2}}+\sqrt[3]{\frac{1}{2}(61+ 9\sqrt{29}})} )$

Hence the answer is $\boxed{yes}$ !

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#3 h Jul 27, 2013, 10:55 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

hello, is there a typo in the given system, i have found an other solution, or maybe i'm wrong?
Sonnhard.

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Dr Sonnhard Graubner

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Dr Sonnhard Graubner

#4 h Jul 27, 2013, 11:07 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

hello, i think it must be
$(x^2+y^2+z^2)^2=13+\frac{12}{5}(x^2+y^2+z^2)$ .
Sonnhard.

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#5 h Jul 27, 2013, 4:09 PM • 2 Y

Y by Adventure10, Mango247

Sonnhard: Yes, your equation is correct, but since we only want to consider when $x, y, z$ are real, we can discard the second root that gives $x^2+y^2+z^2=-\frac{13}{5}$ .

Also, I think lehungvietbao made a mistake in the end when he solved for $x, y, z$ . We have $x^2, y^2, z^2$ as the roots of $f(a)=a^3-5a^2+6a-3$ , but they have to be the three distinct roots, since otherwise Vieta doesn't work. Since $f(a)$ only has one positive root, our answer is $\boxed{\text{no}}$ .

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#6 h Jan 11, 2014, 5:44 PM • 3 Y

Y by triangle112, Adventure10, Mango247

Let $x^2=a, y^2=b,z^2=c, a,b,c>0$ . Now $a^2+b^2+c^2=13, \sqrt{abc}(a+b+c)=5\sqrt{3}, \sqrt{abc}(ab+bc+ca)=6\sqrt{3}$ so $5(ab+bc+ca)=6(a+b+c)$ . We have $a^2+b^2+c^2+\frac{12(a+b+c)}{5}=(a+b+c)^2, 5(a+b+c)^2-12(a+b+c)-65=0$ solving the equation we get $a+b+c=5$ and now it is easy to see that $ab+bc+ac=6,abc=3$ but $36=(ab+bc+ca)^2\ge\ 3abc(a+b+c)=45$ which is not true, so no solution.

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#7 h Dec 16, 2020, 7:45 AM

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Alternative ending:
After
$mnk=3$ $mn+nk+mk=6$ $m+n+k=5$ ( $m=a^2, n=b^2, k=c^2$ )
we get $1/m+1/n+1/k=2$
However, by Holder we have:
$24=6*2^2=(mn+nk+mk)(1/m+1/n+1/k)(1/n+1/k+1/m) \geq (1+1+1)^3=27$ , impossible

This post has been edited 1 time. Last edited by triangle112, Dec 16, 2020, 7:47 AM
Reason: labelsmdfoiwnesfnvw

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