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
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?
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.
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.
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I would only manipulate the LHS since the equality has not yet been proven.
$\displaystyle \frac{b}{c+ta} + \frac{c}{a+tb} = \frac{ab+tb^{2}+c^{2}+tac}{(c+ta)(a+tb)}$
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.