# Thread: Need help in proving an inequality \o\ /o/

1. ## Need help in proving an inequality \o\ /o/

Hello,
I have been working on this problem since ages and I cant seem to prove it could anyone pls help Thank you sooooooooo much

a/b+tc + b/c+ta + c/a+tb >= 3/1+t

with a,b,c > 0 and t>= 0

2. ## Re: Need help in proving an inequality \o\ /o/

Please learn how to use Latex for next time; it makes it more convenient for people who can answer your questions.

$\displaystyle \frac{a}{b+tc} + \frac{b}{c+ta} + \frac{c}{a+tb} \geq \frac{3}{1+t}$

Is this what you mean?

3. ## Re: Need help in proving an inequality \o\ /o/

yes sorry, I didnt realise how to integrate Latex into this forumn ... just type normally with Latex commands ?

4. ## Re: Need help in proving an inequality \o\ /o/

wrap [*tex] and [*/tex] without asterisks around your equation using t^{2} for $\displaystyle t^{2}$ \frac{a}{b} to express the fraction $\displaystyle \frac{a}{b}$ and \leq for less than or equal to or \geq for greater than or equal to. there should be an online table with latex commands which you can study. (usually they also give you an example, in which case copy paste the text and replace numbers/letters as required)

$\displaystyle \frac{a}{b+tc} + \frac{b}{c+ta} + \frac{c}{a+tb} \geq \frac{3}{1+t}$

Without loss of generality, we can order a,b,c such that $\displaystyle a \geq b \geq c \geq 0$ (otherwise relabel)

$\displaystyle \frac{a}{b+tc} \geq \frac{1}{1+t} \iff \frac{b+tc}{a} \leq 1+t$ which is true since $\displaystyle \frac{b}{a} \leq 1$ and $\displaystyle \frac{ct}{a} \leq t$

Now all we are left to show is that

$\displaystyle \frac{b}{c+ta}+\frac{c}{a+tb} \geq \frac{2}{1+t}$ which I'll leave to you. You can't use the same method as I did for the first term since $\displaystyle \frac{at}{b} \geq t$. I would start by expressing the LHS of this inequality with the same denominator.

5. ## Re: Need help in proving an inequality \o\ /o/

With the same deniminator and developing everything and sorting through ts :
$\displaystyle ba + c^2 - 2ac \geq t(2cb + 2a^2 + ab - b^2 - ac - tb - c^2 - tac)$
but i have no idea how to proceed from here

6. ## Re: Need help in proving an inequality \o\ /o/

I was thinking of a more elegant way, and here's what I got. I was close, but $\displaystyle \frac{c}{a} \leq 1$

$\displaystyle \frac{a}{b+tc} + \frac{b}{c+ta} + \frac{c}{a+tb} \geq \frac{c}{b+tc}+\frac{c}{c+ta}+\frac{c}{a+tb}$ by ordering
$\displaystyle \frac{c}{b+tc}+\frac{c}{c+ta}+\frac{c}{a+tb} \geq \frac{c}{a+ta} + \frac{c}{a+ta} + \frac{c}{a+ta} = \frac{c}{a}\frac{3}{1+t}$

since $\displaystyle b+tc \leq a+ta$ + same for others.
--

I would only manipulate the LHS since the equality has not yet been proven.

$\displaystyle \frac{b}{c+ta} + \frac{c}{a+tb} = \frac{ab+tb^{2}+c^{2}+tac}{(c+ta)(a+tb)}$

7. ## Re: Need help in proving an inequality \o\ /o/

I have the proof to a specific case of $\displaystyle t=1$ but it seems that it doesnt work that well in the general case. Used chebyshev .... but it seems that I cannot arrange this one in the same way :/

8. ## Re: Need help in proving an inequality \o\ /o/

I have another question :
Let n> k an integer. Are there any positive integers n such that any k is never
coprime together, but any k + 1 is always coprime.
2) Is there an infinite sequence of positive integers satisfying the two conditions above?

This case .. I didnt not even manage to understand the question and have no idea where to start ... should i post it here ? or start a new thread lol.