This is not original but it is really interesting.

Let \(x\), \(y\) and \(z\) be positive reals such that \(xyz \geq 1\). Prove that

\[\dfrac {x^5 - x^2}{x^5 + y^2 + z^2} + \dfrac {y^5 - y^2}{y^5 + x^2 + z^2}+ \dfrac {z^5 - z^2}{z^5 + x^2 + y^2} \geq 0.\]

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## Comments

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TopNewest@Sharky Kesa I know its not the right place to say this, but the problem of mine you just reshared has been deleted by me because I, by mistake, posted the same problem twice......maybe you'd like to reshare it again....

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I had accidentally reshared it . Using a Samsung Galaxy. Was it the triangle one?

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Yup, the triangle one.......BTW which Samsung Galaxy?

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"probably". Please don't tell me that you're not telling the model of your Galaxy for Cyber-safety............BTW I posted a problem named 'Special Speed', in which we had to find a way of solving a quartic.......and it was hardly 30 seconds that I had posted the problem, and you had solved it......HOW?/! Are you so fast?Log in to reply

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A better way to approach it (another method in my mind) is to factorise it. Upon factorisation, you get \((1 - 2a)^2 (a^2 + 1) \geq 0\). Since \(a\) is real, \((1 - 2a)^2\) and \(a^2 + 1\) result in a non-negative integer.

Hope that helps. :D

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\[(4a^4 + 4a^2) - (4a^3 + 4a) + (a^2 + 1) = (a^2 + 1)(1 - 4a + 4a^2) = (a^2 + 1)(1 - 2a)^2\]

Finding roots are too laborious in certain equations such as this.

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Factorial Funda

Ah...Thanks......BTW see my new problem...Log in to reply

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@Satvik Golechha

Whaat...? Sharky couldnt solve this ? This seems familiar though. I would be thankful If you could provide me the source of this pro-Log in to reply

@Satvik Golechha I love tagging myself..

Whenever I do not mention at the end of the question that it is taken from somewhere, you can assume that the question is mine. And so, the source of this question isLog in to reply

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Bits, Bytes and a GB

Try my new questionLog in to reply

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