Forum de Mathématiques: Maths-Forum

Je tiens à préciser que l'inégalité est par zweig incorect, prendre n = 3.
J'espère que j'ai écrit en français corectly.

Thank you for understanding ,i will expect your indication.
I did not intentionally break the rules,i can't understand French.

Can somebody demonstrate that:

i have another problem and i thought it but without results,if u can help me?
f(x)=???
if:
f(2x+3)-2f(x-2)=26
i try to solve it but i cant,i determined one point:f(-7)=-26,but i need another point to solve the probleme.
i forggot to precise that f had form:ax+b.

ThSQ a écrit:Convexity of ctg + Jensen

tks for answer,i got it

demonstrate that:

acoustica a écrit:Yes, the idea is quite short. It's more difficult to explain with a clear demonstration.

thank you very much for you contribution.a appreciate that ehat you done.
thanks again.

acoustica a écrit:OK :happy2:
Everything is alright?

yes but yest i think that the solution is very short.

Yes. Just a queston: when you write R/Q, it is R but not Q? If it is so, that is i've said: if n is a number of Q, then, you can find p and q of N, so that n=\frac{p}{q} and n has a periodic writing. In romania we say that a number not in Q but in R exemple sqrt(2) this category of number we say th...

acoustica a écrit:I changed slightly what I've written. Is it better? :happy2:

from what u wrote i understand that any number that not have any periodic figure then the number in R/Q.

In french, we say an "universe number", so there is no periodic serie of numbers. In fact, if want to have "54400012646781100078765557", you can, as for every serie of numbers. For the demonstration, you can have how many "0" as you want. :happy2: If i understand what ...

demnostrate that:

*pls try in english.

It means "Arithmetic means - Geometric means", that's : For all positive reals a_1, a_2, \ldots, a_n and positive intenger n , we have : a_1 + a_2 + \cdots + a_n \geq n\cdot\sqrt[n]{a_1a_2\cdots a_n} Yes i know that it's inequality between "arithmetic average" and "geometri...

The inequality to show rewrites like : (ab + ac + bc)(a + b + c) + 3(ab + ac + bc) - 6(a + b + c) \geq 0 From Schur's inequality we have : (ab + ac + bc)(a + b + c) - 9abc \geq 0 \Leftrightarrow ab + ac + bc \geq \frac{9}{a+b+c} Indeed : (ab + ac ...

I may have not have the time right now . but yet you can consider a , b ,c as complex numbers . You may know this equation a² + b² + c² = ab + ac + bc defines an equilateral triangle ABC (a,b,c) in the complex plane . I think you can start from here. yes i try with this method but nothing,it's too ...

Weensie a écrit:All right . Are you familiar with usual inequalities such as Chebychev's , nesbitt ,Holder , Cauchy schwartz etc... ?

only what i want is to solve this problem,i try a lot of but i didnnt succeed
maybe you could.plz solve it if you know to solve it.

can somebody solve this problem:
if

then prove:

plz try to solve it en english becoz i can't understand france
i need it today
thanks