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 :)
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1If you know $x = y = z = \sqrt{2}$, what help do you need? – apnorton Jan 17 '13 at 16:51
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Maybe there are another solution... – Iuli Jan 17 '13 at 16:53
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You can certainly have $\pm\sqrt{2}$ also. – JP McCarthy Jan 17 '13 at 16:55
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It must be a way to prove that $\pm \sqrt{2}$ are the uniqueness – Iuli Jan 17 '13 at 17:04
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Iuli, can you please tell what book you get these from? Interesting questions. – P.K. Jan 17 '13 at 17:22
1 Answers
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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.
Diego
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