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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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.... – Satvik Golechha · 2 years, 8 months ago

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– Sharky Kesa · 2 years, 8 months ago

I had accidentally reshared it . Using a Samsung Galaxy. Was it the triangle one?Log in to reply

– Satvik Golechha · 2 years, 8 months ago

Yup, the triangle one.......BTW which Samsung Galaxy?Log in to reply

– Sharky Kesa · 2 years, 8 months ago

One of them, the latest.Log in to reply

– Satvik Golechha · 2 years, 8 months ago

Lemme guess.......S5 ??Log in to reply

– Sharky Kesa · 2 years, 8 months ago

Probably.Log in to reply

"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 · 2 years, 8 months agoLog in to reply

– Sharky Kesa · 2 years, 8 months ago

Practice and experience is very useful to solve problems quickly and effectively. We just got the Galaxy so I don't know. :DLog in to reply

– Satvik Golechha · 2 years, 8 months ago

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..Log in to reply

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 · 2 years, 8 months ago

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– Satvik Golechha · 2 years, 8 months ago

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?Log in to reply

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

Finding roots are too laborious in certain equations such as this. – Sharky Kesa · 2 years, 8 months ago

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– Krishna Ar · 2 years, 8 months ago

Wow! How can you think of such awesome simplification.... :-o?Log in to reply

– Sharky Kesa · 2 years, 8 months ago

A load of practice and experience.Log in to reply

Factorial Funda – Satvik Golechha · 2 years, 8 months ago

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

– Sharky Kesa · 2 years, 8 months ago

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.Log in to reply

– Satvik Golechha · 2 years, 8 months ago

The question is correct. You're missing something....... :DDLog in to reply

– Sharky Kesa · 2 years, 8 months ago

What? Did I need to include 1?Log in to reply

– Satvik Golechha · 2 years, 8 months ago

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'Log in to reply

@Satvik Golechha – Krishna Ar · 2 years, 8 months ago

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.. – Satvik Golechha · 2 years, 8 months ago

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

– Satvik Golechha · 2 years, 8 months ago

Hahahahahaahahahahahaha I just got a mail....."Satvik Golechha mentioned you on Brilliant."Log in to reply

Bits, Bytes and a GB – Sharky Kesa · 2 years, 8 months ago

Try my new questionLog in to reply

– Satvik Golechha · 2 years, 8 months ago

Was easy.....I solved it.......Though I will really appreciate anybody who solves it without a calc.Log in to reply

– Sharky Kesa · 2 years, 8 months ago

What was your solution?Log in to reply

– Satvik Golechha · 2 years, 8 months ago

I've posted it.....its just above yours. Learnt it in 5th class.Log in to reply