# 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

$\dfrac {x^5 - x^2}{x^5 + y^2 + z^2} + \dfrac {y^5 - y^2}{y^5 + x^2 + z^2}+ \dfrac {z^5 - z^2}{z^5 + x^2 + y^2} \geq 0.$ Note by Sharky Kesa
7 years, 1 month 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....

- 7 years, 1 month ago

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

- 7 years, 1 month ago

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

- 7 years, 1 month ago

One of them, the latest.

- 7 years, 1 month ago

Lemme guess.......S5 ??

- 7 years, 1 month ago

Probably.

- 7 years, 1 month ago

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?

- 7 years, 1 month ago

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

- 7 years, 1 month 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..

- 7 years, 1 month ago

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

- 7 years, 1 month 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?

- 7 years, 1 month ago

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.

- 7 years, 1 month ago

Ah...Thanks......BTW see my new problem...Factorial Funda

- 7 years, 1 month 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.

- 7 years, 1 month ago

The question is correct. You're missing something....... :DD

- 7 years, 1 month ago

What? Did I need to include 1?

- 7 years, 1 month 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'

- 7 years, 1 month ago

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

- 7 years, 1 month 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 is @Satvik Golechha I love tagging myself..

- 7 years, 1 month ago

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

- 7 years, 1 month ago

Try my new question Bits, Bytes and a GB

- 7 years, 1 month ago

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

- 7 years, 1 month ago

- 7 years, 1 month ago

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

- 7 years, 1 month ago

Wow! How can you think of such awesome simplification.... :-o?

- 7 years, 1 month ago

A load of practice and experience.

- 7 years, 1 month ago