If x, y, z are real numbers, show that (y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 >= yz + zx + xy. I simplified it to x^2 + y^2 + z^2 >= xy + xz + yz not really sure what Im supposed to do next though lol
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Multiply both sides by 2: $\displaystyle 2x^2+2y^2+2z^2\geq 2xy+2xz+2yz$ $\displaystyle (x^2-2xy+y^2)+(x^2-2xz+z^2)+(y^2-2yz+z^2)\geq 0$ $\displaystyle (x-y)^2+(x-z)^2+(y-z)^2\geq 0$
That's factorised, but I'm still not sure how I show that the Left Hand Side is >= to the Right Hand Side
$\displaystyle (x-y)^2+(x-z)^2+(y-z)^2\geq 0$ This is true because it is the sum of three positive numbers, which is >=0
Originally Posted by kinetix2006 If x, y, z are real numbers, show that (y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 >= yz + zx + xy. I simplified it to x^2 + y^2 + z^2 >= xy + xz + yz not really sure what Im supposed to do next though lol You can't manipulate the inequality since that's you asked to prove. On red_dog's post, he wrote $\displaystyle (x-y)^2+(x-z)^2+(y-z)^2\geq 0 ,$ which is an obvious fact, start from there to prove what you want.
Originally Posted by Krizalid You can't manipulate the inequality since that's you asked to prove. On red_dog's post, he wrote $\displaystyle (x-y)^2+(x-z)^2+(y-z)^2\geq 0 ,$ which is an obvious fact, start from there to prove what you want. Okay, if you can't manipulate the inequality though, then what would be the procedure for proving it from the starting point of (y + z - x)^2 + (z + x - y)^2 + (x + y - z)^2 >= yz + zx + xy
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