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

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

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– George G · 3 years, 8 months ago

\(\sum_{i=2}^{10}(x^i+y^i+z^i + \frac{1}{9}) = \sum^{10}_{i=2}(x^i+y^i+z^i ) + 1 = \sum^{10}_{i=1}(x^i+y^i+z^i)\)Log in to reply

– Michael Tong · 3 years, 8 months ago

sorry, brainfart, failed to realize that the \(x + y + z\) is also multiplied by \(\frac{1}{10}\), silly me.Log in to reply

– Piyushkumar Palan · 3 years, 8 months ago

yeah.. precisely!Log in to reply

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 · 3 years, 6 months ago

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