\(x, y, z\)

Let x,y,zx, y, z be three positive integers such that x+y+z=1x+y+z=1. Show that x5x+y5y+z5z110i=210(xi+yi+zi+19)x^5\sqrt{x}+y^5\sqrt{y}+z^5\sqrt{z}\leq\frac{1}{10}\sum\limits_{i=2}^{10}\left(x^i+y^i+z^i+\frac{1}{9}\right )

#Algebra #MathProblem #Math

Note by Jaydee Lucero
7 years, 6 months ago

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9 votes

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Comments

x,y ,z positive reals?

It follows from AM-GM: i=110xi10x5x\sum_{i=1}^{10} x^i \ge 10 x^5\sqrt{x}. Then just add three together.

George G - 7 years, 6 months ago

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yeah.. precisely!

Piyushkumar Palan - 7 years, 6 months ago

if the base of the isosceles triangle joints the points (2,-4), (1,-3); the area is 9/2 . find the third vertex of the triangle.

please help me, show your solution

Jeriel Villa - 7 years, 4 months ago
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