Problem B. 5335. (October 2023)
B. 5335. The product of the positive numbers \(\displaystyle x\), \(\displaystyle y\), \(\displaystyle z\) is 1. What may be the value of the expression
\(\displaystyle \left(x+\frac{1}{x}\right)^{2} +\left(y+\frac{1}{y}\right)^{2} +\left(z+\frac{1}{z}\right)^{2} -\left(x+\frac{1}{x}\right) \left(y+\frac{1}{y}\right) \left(z+\frac{1}{z}\right)? \)
Proposed by G. Kiss, Csömör
(3 pont)
Deadline expired on November 10, 2023.
Sorry, the solution is available only in Hungarian. Google translation
Megoldás. Felbontva a zárójeleket, majd kihasználva, hogy az \(\displaystyle xyz = 1\) feltétel szerint \(\displaystyle \frac{xy}{z} = \frac1{z^2}\), \(\displaystyle \frac{xz}{y} = \frac1{y^2}\), \(\displaystyle \frac{yz}{x} = \frac1{x^2}\), \(\displaystyle \frac{x}{yz} = x^2\), \(\displaystyle \frac{y}{xz} = y^2\), \(\displaystyle \frac{z}{xy} = z^2\) és \(\displaystyle \frac1{xyz} = 1\), a következőt kapjuk:
$$\begin{eqnarray*} & & \left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(z+\frac{1}{z}\right)^2-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(z+\frac{1}{z}\right) = \\ & = & x^2 + 2 + \frac1{x^2} + y^2 + 2 + \frac1{y^2} + z^2 + 2 + \frac1{z^2} - xyz - \frac{xy}{z} - \frac{xz}{y} - \frac{yz}{x} - \frac{x}{yz} - \frac{y}{xz} - \frac{z}{xy} - \frac1{xyz} = \\ & = & x^2 + 2 + \frac1{x^2} + y^2 + 2 + \frac1{y^2} + z^2 + 2 + \frac1{z^2} - xyz - 1 - \frac1{z^2} - \frac1{y^2} - \frac1{x^2} - x^2 - y^2 - z^2 - 1 = 4. \end{eqnarray*}$$Tehát a kifejezés értéke mindig 4 (ha az \(\displaystyle x,y,z\) számhármas szorzata 1).
Statistics:
223 students sent a solution. 3 points: 173 students. 2 points: 15 students. 1 point: 1 student. 0 point: 6 students. Not shown because of missing birth date or parental permission: 20 solutions.
Problems in Mathematics of KöMaL, October 2023