Solve in the positive real numbers, the system of equations : $$(2x)^{2013} + (2y)^{2013} + (2z)^{2013} =3 $$ and $$xy + yz + zx + 2xyz = 1.$$
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Thanks for the edit , @julien but can you tell me how did you break the lines? – Kushashwa Ravi Shrimali Mar 22 '13 at 11:03
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apart from (1/2, 1/2, 1/2), I imagine :-) – mau Mar 22 '13 at 11:03
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To get an equation centered on a single line, put it between double dollar sign, instead of single dollar. – Julien Mar 22 '13 at 11:05
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Ok! Thanks @julien I would take care of it from now. – Kushashwa Ravi Shrimali Mar 22 '13 at 11:07
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1@julien, I'm about to remove the differential-geometry tag, since Ivan's solution makes it clear that that subject doesn't enter into it. – Gerry Myerson Mar 22 '13 at 12:04
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Let $x=\frac{a}{2}, y=\frac{b}{2}, z=\frac{c}{2}$, so
$$ab+bc+ca+abc=4, a^{2013}+b^{2013}+c^{2013}=3$$
Now by A.M.$\geq$G.M. $4=ab+bc+ca+abc \geq 3\sqrt[3]{a^2b^2c^2}+abc$. Let $x=\sqrt[3]{abc}$, so that $4 \geq 3x^2+x^3$.
Factorising, we get $(x-1)(x+2)^2 \leq 0$, so $x \leq 1$. Thus $abc \leq 1$, so $ab+ac+bc=4-abc \geq 3$.
Now by power mean inequality and the inequality $a^2+b^2+c^2 \geq ab+ac+bc$ (which is equivalent to $\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2) \geq 0$), we get
$$1=\sqrt[2013]{\frac{a^{2013}+b^{2013}+c^{2013}}{3}} \geq \sqrt{\frac{a^2+b^2+c^2}{3}} \geq \sqrt{\frac{ab+bc+ca}{3}} \geq 1$$
Since equality holds, we must have $a=b=c$, so $3=3a^{2013}$ and we get $a=b=c=1$.
Therefore the only solution in positive reals is $(x, y, z)=(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$.
Ivan Loh
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