100 got mad!

Algebra Level 4

$\begin{cases} xyzw & = 1 \\ x^{100} + 100y & = 100z + w^{100} \\ 100x + y^{100} & = z^{100} + 100w \end{cases}$

Let $(x,y,z,w)$ be positive real numbers that satisfy the system of equations above. Find the minimum possible value of $x + y + z + w$ .

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The answer is 4.

1 solution

Steven Jim
Jun 1, 2017

As $x, y, z, w$ are positive real numbers, we have that $\frac { x+y+z+w }{ 4 } \ge \sqrt [ 4 ]{ xyzw } =1$ . Thus $min(x + y + z + w) = 4$ .

By the way, the second and third equation is somewhat useless.

So my question to you is the same question with same conditions as above, but instead of second equation you have (x^100)+101y = 100z +(w^100) And now think over this and tell me the minimum value of x+y+z+w

Ayush Sharma - 4 years ago

That is why I said that it's pretty much useless when the second and third equation also holds true when x = y = z = w = 1.

Steven Jim - 4 years ago

ya brother thats right the equations are not useless instead they hold true by default