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This is not original but it is really interesting.

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

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

Note by Sharky Kesa
3 years ago

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@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.... Satvik Golechha · 3 years ago

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@Satvik Golechha I had accidentally reshared it . Using a Samsung Galaxy. Was it the triangle one? Sharky Kesa · 3 years ago

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@Sharky Kesa Yup, the triangle one.......BTW which Samsung Galaxy? Satvik Golechha · 3 years ago

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@Satvik Golechha One of them, the latest. Sharky Kesa · 3 years ago

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@Sharky Kesa Lemme guess.......S5 ?? Satvik Golechha · 3 years ago

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@Satvik Golechha Probably. Sharky Kesa · 3 years ago

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@Sharky Kesa What d'you mean by "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? Satvik Golechha · 3 years ago

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@Satvik Golechha Practice and experience is very useful to solve problems quickly and effectively. We just got the Galaxy so I don't know. :D Sharky Kesa · 3 years ago

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@Sharky Kesa Will you please help me in some questions.....I'm callow to inequalities. For example, For any real number 'a', prove that \(\normalsize 4a^4-4a^3+5a^2-4a+1\ge0\). I want to know what was the first thing that came to your mind when you saw this question......and then how you approached it. Thanks in advance.. Satvik Golechha · 3 years ago

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@Satvik Golechha When \(a\) is positive, the whole thing results in a non-negative numeral (\(\frac {1}{2}\) is the equality case). When \(a\) is negative, the odd powers are in subtraction so the minus' turn to plus'. They also result in positive numbers. When \(a\) is 0, the whole thing results in positive. That's one of the ways to approach it.

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 Sharky Kesa · 3 years ago

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@Sharky Kesa Thanks......I could do it by the first method myself. I wanted to use the second method, but I couldn't factorize it. Did you factorize it by finding out the roots? If so, then what if it didn't have any real root? Satvik Golechha · 3 years ago

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@Satvik Golechha Nah, I simply rearranged the equation to this:

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

Finding roots are too laborious in certain equations such as this. Sharky Kesa · 3 years ago

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@Sharky Kesa Wow! How can you think of such awesome simplification.... :-o? Krishna Ar · 3 years ago

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@Krishna Ar A load of practice and experience. Sharky Kesa · 3 years ago

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@Sharky Kesa Ah...Thanks......BTW see my new problem...Factorial Funda Satvik Golechha · 3 years ago

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@Satvik Golechha You will not believe this! I first attempted the number 168 (number of primes under 1000, hence the number of numbers for which \(n\) can not divide \((n-1)!\). Then, I relied what I did wrong and did 832. It was wrong, and I thought I was including 1 and 1000 and finally did 830. You know the answer and I couldn't believe it. Sharky Kesa · 3 years ago

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@Sharky Kesa The question is correct. You're missing something....... :DD Satvik Golechha · 3 years ago

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@Satvik Golechha What? Did I need to include 1? Sharky Kesa · 3 years ago

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@Sharky Kesa Since you've had all your tries, I can tell you the solution......but not here. What's your mail id?......If you don't want to give it here, you may send me a test mail at 'sgsuper@yahoo.com' Satvik Golechha · 3 years ago

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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- @Satvik Golechha Krishna Ar · 3 years ago

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@Krishna Ar 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 is @Satvik Golechha I love tagging myself.. Satvik Golechha · 3 years ago

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@Satvik Golechha Hahahahahaahahahahahaha I just got a mail....."Satvik Golechha mentioned you on Brilliant." Satvik Golechha · 3 years ago

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@Satvik Golechha Try my new question Bits, Bytes and a GB Sharky Kesa · 3 years ago

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@Sharky Kesa Was easy.....I solved it.......Though I will really appreciate anybody who solves it without a calc. Satvik Golechha · 3 years ago

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@Satvik Golechha What was your solution? Sharky Kesa · 3 years ago

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@Sharky Kesa I've posted it.....its just above yours. Learnt it in 5th class. Satvik Golechha · 3 years ago

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