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# Algebra

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163
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find the minimum possible value of $$\frac{(y-x)^2}{(y-z)(z-x)}+\frac{(z-y)^2}{(z-x)(x-y)}+\frac{(x-z)^2}{(x-y)(y-z)},$$ where x, y, and z are distinct real numbers

the answer is 3 but why?

Apr 18, 2022
edited by Guest  Apr 18, 2022

#1
+115
-3

Hmm let me see if i can see what your problem

Apr 18, 2022
#2
+9461
+1

Note that

$$\begin{array}{cl} &\displaystyle\frac{(y-x)^2}{(y-z)(z-x)}+\frac{(z-y)^2}{(z-x)(x-y)}+\frac{(x-z)^2}{(x-y)(y-z)} \\ =& \dfrac{-(y - x)^3-(z - y)^3 -(x - z)^3}{(y - z)(z - x)(x - y)}\\ =& \dfrac{(x - y)^3 + (y - z)^3 + (z - x)^3}{(y - z)(z - x)(x - y)} \end{array}$$

Now, we use the identity $$a^3 + b^3 + c^3 - 3abc = (a + b +c)(a^2 + b^2 + c^2 - ab - bc -ca)$$, with $$a = x - y, b =y - z, c = z - x$$

(If you don't believe me, you can try to expand the right-hand side of the identity. It should take some effort, but you will eventually arrive at the left-hand side.)

$$\begin{array}{rcl} (x - y)^3 + (y - z)^3 + (z - x)^3 - 3(x - y)(y - z)(z - x) &=& ((x - y) + (y - z) + (z - x))(\text{something})\\ &=& (x - y + y - z + z - x)(\text{something})\\ &=& (0)(\text{something})\\ &=& 0\\ (x - y)^3 + (y - z)^3 + (z - x)^3 &=& 3(x - y)(y - z)(z - x) \end{array}$$

Therefore, the numerator is exactly 3 times the denominator.

We have, for any distinct real numbers x, y, z, $$\displaystyle\frac{(y-x)^2}{(y-z)(z-x)}+\frac{(z-y)^2}{(z-x)(x-y)}+\frac{(x-z)^2}{(x-y)(y-z)} = 3$$.

Since it is constantly 3, the minimum value must be 3.

Apr 19, 2022