By Cauchy Schwarz inequality: $ (x+y+z+t) \cdot (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}) \geq (1+1+1+1)^2=16 $ So, $4 \cdot (5-\frac{1}{xyzt}) \geq 16$ $5-\frac{1}{xyzt} \geq 4$ $\frac{1}{xyzt} \le 1$ $xyzt \geq 1$ Now, by AM-GM: $x+y+z+t \geq 4 \cdot \sqrt[4]{xyzt} $ $\sqrt[4]{xyzt} \le 1$ $xyzt \le 1$ $1 \le xyzt \le 1$ So, $xyzt=1$, and we have an equality, and thats iff $x=y=z=t=1$. Solution is(indeed) $(x,y,z,t)=(1,1,1,1)$

I'm not sure i understood how you came to the conclusion that $x=y=z=t$. In general there can be other solutions not obeying this restriction. we should build the polynomial of the roots - here it will be $t^{4} - 4t^{3}+ 6t^{2} - 4t +1 = 0$ which can be solved quite easily due to the symmetric coefficients. then indeed you have only $x=y=z=t=1$.

yair wrote: I'm not sure i understood how you came to the conclusion that $x=y=z=t$. Both Cauchy-Schwarz and AM-GM have as equality cases $x=y=z=t$. yair wrote: In general there can be other solutions not obeying this restriction. Which ones? yair wrote: We should build the polynomial of the roots - here it will be $t^{4} - 4t^{3}+ 6t^{2} - 4t +1 = 0$ which can be solved quite easily due to the symmetric coefficients. then indeed you have only $x=y=z=t=1$. And how do you compute the coefficients $-4,6,-4,1$ ? Only the first $-4$ is immediate, from $x+y+z+t=4$.

yes you're right, it came from c-s... when i wrote In general there can be other solutions not obeying this restriction I meant in other case (not this specific system). still, the calculation of the polynomial is simple. except $x+y+z+t=4$ and $xyzt=1$ which are immediate we can form another one using the second equation of the given system and fourth one using the condition that one root is 1.

It was the 1st problem of Greek Team Selection Test 2010.

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\ge 4=>xyzt\ge 1$ $x+y+z+t=4=>xyzt\le 1=>xyzt=1$ and the case of equality holds if and only if when $x=y=z=t=>x=y=z=t=1$