Maximum or Minimum?

Algebra Level 4

x y + y z + z x x 2 + y 2 + z 2 \large \dfrac{xy + yz + zx}{x^2 + y^2 + z^2}

Given that x , y x,y and z z are real numbers. Which of the following can't be equal to above expression?

Original Problem.
0 0 1 1 − 1 2 -\dfrac{1}{2} 1 2 \dfrac{1}{2} 2 2

Akhil Bansal by Akhil Bansal , 22, India

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1 solution

Akhil Bansal
Nov 30, 2015

For maximum,
Not that ( x − y ) 2 + ( y − z ) 2 + ( z − x ) 2 ≥ 0 (x-y)^2 + (y-z)^2 + (z-x)^2 \geq 0 . Then
⇒ 2 [ x 2 + y 2 + z 2 − ( x y + y z + z x ) ] ≥ 0 \large \Rightarrow 2 \left[x^2 + y^2 +z^2 - (xy +yz +zx) \right] \geq 0 ⇒ x 2 + y 2 + z 2 ≥ x y + y z + z x \large \Rightarrow x^2 + y^2 +z^2 \geq xy +yz +zx ⇒ x y + y z + z x x 2 + y 2 + z 2 ≤ 1 \large \Rightarrow \dfrac{xy +yz +zx}{x^2 + y^2 +z^2} \leq 1

For minimum,

Note that ( x + y + z ) 2 ≥ 0 (x + y + z)^2 \ge 0 . Then x 2 + y 2 + z 2 + 2 x y + 2 x z + 2 y z ≥ 0 x^2 + y^2 + z^2 + 2xy + 2xz + 2yz \geq 0 ⇒ 2 ( x y + x z + y z ) ≥ − ( x 2 + y 2 + z 2 ) \Rightarrow 2(xy + xz + yz) \geq -(x^2 + y^2 + z^2) ⇒ x y + x z + y z x 2 + y 2 + z 2 ≥ − 1 2 \Rightarrow \dfrac{xy + xz + yz}{x^2 + y^2 + z^2} \geq -\frac{1}{2}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

For completeness, you should show that the other values can indeed be attained. Had you done that, you would have realized that we could not obtain -1 or -2.

(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) x^2+y^2+z^2>=-2(xy+yz+zx) So (xy+yz+zx)/(x^2+y^2+z^2)>=(-1/2) So answer should be even -2

Pranav Rao - 5 years, 6 months ago

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The given expression is ( x + y + z ) 2 2 ( x 2 + y 2 + z 2 ) − 1 2 ≥ − 1 2 \frac{(x+y+z)^2}{2(x^2+y^2+z^2)}-\frac{1}{2}\geq -\frac{1}{2} , so that the value cannot be 2 , − 1 2,-1 or − 2 -2 . This problem is flawed.

Otto Bretscher - 5 years, 6 months ago

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Yes, I got that Sir. Can you please report it..

Akhil Bansal - 5 years, 6 months ago

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