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 $\boxed{yes}$ !

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

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

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}} $.

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.

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