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Lê Quốc Trần Anh Coordinator

07/05/2018 at 11:41
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Show geometrically (like in my solution) that for positive reals $x,\,y,\,z$,
\[xy+yz+zx\leq x^2+y^2+z^2.\]



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    Dao Trong Luan Coordinator 10/05/2018 at 11:55

    Multiplying each side with 2, we have:

    2xy + 2yz + 2zx ≤ 2x2 + 2y2 + 2z2

    <=> 2xy + 2yz + 2zx - 2x2 - 2y2 - 2z2 ≤ 0

    <=> 2xy + 2yz + 2zx - x2 - x2 - y2 - y2 - z2 - z2 ≤ 0

    <=> -(x2 - 2xy + y2) - (x2 - 2xz + z2) - (y2 + 2yz + z2) ≤ 0

    <=> -(x+y)2 - (x+z)2 - (y+z)2 ≤ 0 (True)     

    So xy + yz + zx ≤ x2 + y2 + z2

    Lê Quốc Trần Anh selected this answer.


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