SOLUTION: Let be three positive numbers such that: x^2 + y^2 + z^2 = 2(xy + xz + yz). x + y + z + \frac{1}{xyz} > 4.

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Question 1209502: Let be three positive numbers such that:
x^2 + y^2 + z^2 = 2(xy + xz + yz).
x + y + z + \frac{1}{xyz} > 4.

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Certainly, let's prove the given inequality.
**1. Utilize the Given Condition**
* We are given: x² + y² + z² = 2(xy + xz + yz)
* Rearrange the equation:
x² + y² + z² - 2xy - 2xz - 2yz = 0
* Factor the left-hand side:
(x - y)² + (y - z)² + (z - x)² = 0
* Since the squares of real numbers are always non-negative, the only way for the sum of three squares to be zero is if each individual square is zero.
* Therefore:
x - y = 0
y - z = 0
z - x = 0
* This implies that:
x = y = z
**2. Substitute and Simplify**
* Let x = y = z = k (where k is a positive number)
* The inequality to prove becomes:
k + k + k + 1/(k * k * k) > 4
3k + 1/k³ > 4
**3. Prove the Inequality**
* **Method 1: AM-GM Inequality**
* Apply the AM-GM inequality to the numbers 3k and 1/k³:
(3k + 1/k³) / 2 ≥ √(3k * 1/k³)
(3k + 1/k³) / 2 ≥ √3
3k + 1/k³ ≥ 2√3
* Since 2√3 > 4, we have:
3k + 1/k³ > 4
* **Method 2: Algebraic Manipulation**
* Multiply both sides of the inequality by k³ (since k is positive, this does not change the direction of the inequality):
3k⁴ + 1 > 4k³
* Rearrange:
3k⁴ - 4k³ + 1 > 0
* Factor:
(k - 1)³ (3k + 1) > 0
* Since k is positive, (3k + 1) is always positive.
* Therefore, for the inequality to hold, (k - 1)³ must be positive. This implies k > 1.
* Since x = y = z = k, we have x > 1, y > 1, and z > 1.
* Now, consider the expression:
x + y + z + 1/(xyz)
* Since x, y, and z are all greater than 1, 1/(xyz) will be less than 1.
* Therefore, x + y + z + 1/(xyz) > x + y + z > 3 > 4
**Conclusion**
We have proven that if x², y², and z² satisfy the given condition, then x + y + z + 1/(xyz) > 4.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Let x, y, z be three positive numbers such that:
x^2 + y^2 + z^2 = 2(xy + xz + yz).
Prove that x + y + z + 1/(xyz) > 4.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


                The solution in the post by  @CPhill is   .


        Indeed,  he rearranges the given equation   x^2 + y^2 + z^2 = 2(xy + xz + yz)

        to equation   x^2 + y^2 + z^2 - 2xy - 2xz - 2yz = 0,   which is right.


        But then he  " factors "  the left-hand side

                (x-y)^2 + (y-z)^2 + (z-x)^2 = 0,

        which is  FATALLY  WRONG.


        Hence,  everything what  @CPhill derives further from the  " factored "  form is incorrect
        and does not have the power of logical deducing.

                 +-----------------------------------------------------------------+
                    Nevertheless, the statement of the problem is correct,
                                and below I developed another solution.
                 +-----------------------------------------------------------------+ 


We start from equation

    x^2 + y^2 + z^2 = 2xy + 2xz + 2yz,    (1)


which is given.   Add 2xy + 2xz + 2yz to both side.  You will get

    x^2 + y^ + z^2 + 2xy + 2xz + 2yz = 4xy + 4xz + 4 yz.    (2)


Left side of the last equation is  (x+y+z)^2,  so it can be re-written in the form

    (x + y + z)^2 = 4(xy + xz + yz).    (3)


Now apply AM-GM (Arithmetic mean - Geometric mean) inequality, which says

     >=     (4)

for any three positive numbers a, b, c.

Therefore, for the right side of equation (3) we have this inequality

    4(xy + xz + yz) =  >=  = .    (5)


So, combining (3) and (5), we get

    (x + y + z)^2 >= .    (6)


Now take square root of both sides of (6) and get

    x + y + z >= .     (7)


Now, to prove the inequality as it is given in the condition, add  to both sides of (7).
You will get

    x + y + z +  >=  + .    (8)


Consider function   + .


This function is easy to analyze for its minimum.

Such an analysis can be done using standard Calculus procedure or graphically.

The plot of this function is shown in Figure under this link

https://www.desmos.com/calculator/zw8xlstrvb


The minimum is  4.45566.  For us, it gives the information that right side of (8) is always greater than 4.


So, the inequality

    x + y + z +  > 4

is proved.

At this point, the solution to the problem is complete.



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