It can be done, I just need to get a breakthrough

This is not original but it is really interesting.

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

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

#Algebra #Sharky

Easy Math Editor

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Comments

@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 - 6 years, 11 months ago

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

Sharky Kesa - 6 years, 11 months ago

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

Satvik Golechha - 6 years, 11 months ago

@Satvik Golechha – One of them, the latest.

Sharky Kesa - 6 years, 11 months ago

@Sharky Kesa – Lemme guess.......S5 ??

Satvik Golechha - 6 years, 11 months ago

@Satvik Golechha – Probably.

Sharky Kesa - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@Sharky Kesa – Ah...Thanks......BTW see my new problem...Factorial Funda

Satvik Golechha - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@Sharky Kesa – The question is correct. You're missing something....... :DD

Satvik Golechha - 6 years, 11 months ago

@Satvik Golechha – What? Did I need to include 1?

Sharky Kesa - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@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 - 6 years, 11 months ago

@Satvik Golechha – Hahahahahaahahahahahaha I just got a mail....."Satvik Golechha mentioned you on Brilliant."

Satvik Golechha - 6 years, 11 months ago

@Satvik Golechha – Try my new question Bits, Bytes and a GB

Sharky Kesa - 6 years, 11 months ago

@Sharky Kesa – Was easy.....I solved it.......Though I will really appreciate anybody who solves it without a calc.

Satvik Golechha - 6 years, 11 months ago

@Satvik Golechha – What was your solution?

Sharky Kesa - 6 years, 11 months ago

@Sharky Kesa – I've posted it.....its just above yours. Learnt it in 5th class.

Satvik Golechha - 6 years, 11 months ago

@Sharky Kesa – Wow! How can you think of such awesome simplification.... :-o?

Krishna Ar - 6 years, 11 months ago

@Krishna Ar – A load of practice and experience.

Sharky Kesa - 6 years, 11 months ago