<s>[INST] Find $x,y,z$ if $3^{x^4+y^2}+3^{y^4+z^2}+3^{z^4+x^2}=3^7$. How can I solve the following system with $x,y,z$ real numbers:

$$x^2+y^2+z^2=6$$
  $$3^{x^4+y^2}+3^{y^4+z^2}+3^{z^4+x^2}=3^7.$$

I observe that $x=y=z=\sqrt{2}$ and I feel it must apply the inequality $AM \geq GM$ but I don't know how to do.
Thanks :) 
 [/INST] Applying $AM\geq GM$ to your second equation you get
$$3^6\geq ({3^{x^4+y^2+y^4+z^2+z^4+x^2}})^{1/3}=(3^{x^4+y^4+z^4})^{1/3}\cdot(3^{x^2+y^2+z^2})^{1/3}=(3^{x^4+y^4+z^4})^{1/3}\cdot 3^2$$
and therefore $$3^4\geq 3^{\frac{x^4+y^4+z^4}{3}}$$
Now we know that $\sqrt{\frac{x^4+y^4+z^4}{3}}\geq \frac{x^2+y^2+z^2}{3}=2$ where the equality holds if and only if $x^2=y^2=z^2$. This implies that  $$3^4\geq 3^{\frac{x^4+y^4+z^4}{3}}\geq 3^4.$$
Then we must have $\sqrt{\frac{x^4+y^4+z^4}{3}}=\frac{x^2+y^2+z^2}{3}$ and therefore $x^2=y^2=z^2=2$. Then this is your unique solution.
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