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# $$x, y, z$$

Let $$x, y, z$$ be three positive integers such that $$x+y+z=1$$. Show that $x^5\sqrt{x}+y^5\sqrt{y}+z^5\sqrt{z}\leq\frac{1}{10}\sum\limits_{i=2}^{10}\left(x^i+y^i+z^i+\frac{1}{9}\right )$

Note by Jaydee Lucero
3 years, 8 months ago

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x,y ,z positive reals?

It follows from AM-GM: $$\sum_{i=1}^{10} x^i \ge 10 x^5\sqrt{x}$$. Then just add three together. · 3 years, 8 months ago

Comment deleted Dec 12, 2013

$$\sum_{i=2}^{10}(x^i+y^i+z^i + \frac{1}{9}) = \sum^{10}_{i=2}(x^i+y^i+z^i ) + 1 = \sum^{10}_{i=1}(x^i+y^i+z^i)$$ · 3 years, 8 months ago

sorry, brainfart, failed to realize that the $$x + y + z$$ is also multiplied by $$\frac{1}{10}$$, silly me. · 3 years, 8 months ago

yeah.. precisely! · 3 years, 8 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.