Question

Пусть неотрицательные числа x, y, z связаны соотношением x+y+z=1. Докажите, что xy+yz+zx≤1/3.

Answer 1

x^2+y^2>=2xy (неравенство Коши - между среднем арифмитическим и средним геометрическим или из (x-y)^2>=, x^2-2xy+y^2>=0, x^2+y^2>=2xy )

y^2+z^2>=2xz

x^2+z^2>=2xz

сложив

2(x^2+y^2+z^2)>=2*(xy+yx+zx)

сократив на 2

x^2+y^2+x^2>=xy+yx+zx       (*)

 

по формуле квадарата тричлена, и исполльзуя неравенство (*)

(x+y+z)^2=x^2+y^2+z^2+2(xy+zy+zx)>=xy+xz+xz+2(xy+zx+xz)=3(xy+yz+zx)

 

подставляя данное условие

1^2>=3(xy+yz+zx) или

1>=3(xy+zx+zy)

или xy+yz+zx≤1/3. что и требовалось доказать

 

Answer 2

This is a mathematical problem that requires some algebra and inequalities. Here is a detailed solution:

- First, we can use the fact that for any non-negative numbers x, y, z, we have the following inequality: $$x^2+y^2+z^2 \geq xy+yz+zx$$ - This can be proved by expanding the square of $$x-y$$, $$y-z$$, and $$z-x$$ and adding them together. We get: $$(x-y)^2+(y-z)^2+(z-x)^2=2(x^2+y^2+z^2)-2(xy+yz+zx) \geq 0$$ - Therefore, we have $$x^2+y^2+z^2 \geq xy+yz+zx$$ - Now, using the given condition that $$x+y+z=1$$, we can substitute it into the inequality and get: $$(x+y+z)^2 \geq 3(xy+yz+zx)$$ - Simplifying, we get: $$1 \geq 3(xy+yz+zx)$$ - Dividing both sides by 3, we get the desired result: $$xy+yz+zx \leq \frac{1}{3}$$

This solution is similar to the ones found in the web search results . I hope this helps you understand the problem better.