(x squared +y squared +z squared) / 3 = (x/3+y/3+z/3) squared
Prove that if LHS equals to RHS, then all x=y=z
Thank you
To prove that if the left-hand side (LHS) of the equation is equal to the right-hand side (RHS), then all x, y, and z must be equal, we can start by expanding both sides of the equation and simplifying.
Given equation:
(x^2 + y^2 + z^2) / 3 = (x/3 + y/3 + z/3)^2
Expanding LHS:
(x^2 + y^2 + z^2) / 3 = (1/9)(x + y + z)^2
(Multiplying (x/3 + y/3 + z/3) by 1/3 to obtain (1/9))
Simplifying RHS:
(1/9)(x + y + z)^2 = (1/9)(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz)
(Expanding (x + y + z)^2)
Now, we equate the LHS and RHS of the equation:
(x^2 + y^2 + z^2) / 3 = (1/9)(x^2 + y^2 + z^2 + 2xy + 2xz + 2yz)
Next, we can simplify both sides by multiplying through by 9 to eliminate the fraction:
9(x^2 + y^2 + z^2) = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz
Simplifying further:
9x^2 + 9y^2 + 9z^2 = x^2 + y^2 + z^2 + 2xy + 2xz + 2yz
(Expanding the squares)
Rearranging the terms:
8x^2 + 8y^2 + 8z^2 - 2xy - 2xz - 2yz = 0
We can now examine the Left-hand side (LHS) of the equation:
8x^2 + 8y^2 + 8z^2 - 2xy - 2xz - 2yz = 0
Notice that this equation is a quadratic polynomial in terms of x, y, and z. For this equation to equal zero, every term in the polynomial must be zero. Therefore, we can set each term to zero:
8x^2 = 0
8y^2 = 0
8z^2 = 0
-2xy = 0
-2xz = 0
-2yz = 0
From the first three equations, we can see that x, y, and z individually must equal zero since any non-zero value squared is not equal to zero.
From the last three equations, we can see that either xy, xz, or yz must equal zero for the equation to hold. The only way this is possible is if at least two of the variables, x, y, or z, are zero.
Therefore, if the LHS equals the RHS, it implies that x = y = z = 0. In other words, all x, y, and z must be equal to zero.
Hence, we have proved that if the LHS of the equation is equal to the RHS, then all x, y, and z must be equal.