$$(\dfrac{x^7+y^7+1}{\sqrt{8}(x+y)})^2 > \dfrac{x^3y^3}{x^2+y^2+2}$$
Edited by Oral1020, 29-01-2013 - 11:41.
$(\dfrac{x^7+y^7+1}{\sqrt{8}(x+y)})^2 > \dfrac{x^3y^3}{x^2+y^2+2}$
Started By Oral1020, 29-01-2013 - 10:53
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Posted 29-01-2013 - 10:53
Oral1020
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Cho $x,y$ là các số thực dương.Chứng minh rằng:
$$(\dfrac{x^7+y^7+1}{\sqrt{8}(x+y)})^2 > \dfrac{x^3y^3}{x^2+y^2+2}$$
$$(\dfrac{x^7+y^7+1}{\sqrt{8}(x+y)})^2 > \dfrac{x^3y^3}{x^2+y^2+2}$$
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