triclino

prove that the following sequence converges: [math]x_{k+2} =\frac{2}{x_{k+1} + x_{k}} ,x_{1},x_{2}>0[/math]

So if somebody is sick and needs help he/she must do not go to the Doctor but try to fix him/her self

Mooeypoo if i was to ask your experts, for example, for a formal proof of a high school question they would loose their spoon . There are a lot of things that your experts do not know. If for example i was to ask your experts for a formal proof in Analysis probably i would wait centuries for an answer . Boasting around is not a good thing. I can even ask questions in Analysis (not formal ) that your experts wont know the answer. So if you are so sure about your experts let us try . I am not ordering anybody around.

Yes that substitution led us to infinite solutions

And if he was asked to prove it ,then he would realize that it is an impossible task Can you prove it??

inequalities like this are countless .The question is, is there a general way to tackle them??

Solve the following inequality: [math] xy+(xy)^2 +(xy)^3>14[/math] needless to say i have no idea how even to start this inequality

So i take it that you cannot formally analyze your own proof??

If i were to prove that your proof is not correct we will have to formally analyze it But can you formally analyze your own proof??

For a start ,if i were to say that your method 2 is wrong what would you say??

.................................WRONG..................................................... This proof is completely wrong .Go over your proof again .Ask a teacher a Doctor ,anyone . This proof is wrong

O.k teacher here we go: case 1 : ab>0 ,then -ab<0 and since [math]0\leq a^2+b^2[/math] we have: [math] -ab\leq a^2+b^2[/math] case 2 : ab=0 ,then -ab=0 and since [math]0\leq a^2+b^2[/math] we have: [math] -ab\leq a^2+b^2[/math] case 3 : ab<0 ,then -ab>0. And here is the problem. Now we cannot say : [math] -ab\leq a^2+b^2[/math]. Because if you have two Nos x>0 and y>0 you cannot say x>y

No, No you curry on and finish the proof because i want stop posting in this forum

D.H, why you throw your problems on me .I think this is unfair. Besides i accept advice and teaching from anybody as long as it is correct.

And you should learn to read the whole thread before you get involved in a discussion and come into worthless and impolite conclusions If you prove the latter i will stop posting in this forum. The latter is unprovable (unless you do the same mistake with D.H to use [math]a^2+ab+b^2\geq 0[/math] in your proof)

I suggest you go and read (perhaps in a book of logic) what is the double implication proof .Check the following example: 0x = 0 <=> 0x+x = 0+x <=> x(0+1) = 0 +x <=> x= 0+x <=> x=x We want to prove 0x = 0 (for all real ,x) ,so by double implication we arrive at a well known fact x=x ,then we can except 0x=0 as true. So to go from [math]a^2+ab+b^2\geq 0[/math] to [math] (a+b)^2\geq ab[/math] you do the following: [math]a^2+ab+b^2\geq 0\Longleftrightarrow a^2+ab+b^2+ab\geq ab\Longleftrightarrow (a+b)^2\geq ab[/math] Is [math](a+b)^2\geq ab[/math] a well known fact??

In post #7 i proved the inequality in concern .Read it if you like. The hole proof is just a line . I would like to comment on your proof ,but as usually the moderators will intervene and give me valuable advice on how to behave e.t.c Now, on your comment that i have problems with proofs i can say the following: To really say whether a proof is correct or not you must analyze the proof by writing a formal issue of it. Then i will except you as capable of criticizing my proofs. Can you write a formal proof of any high school theorem?

Now i can see you have problems with inequalities. In post # 9 i showed exactly the opposite. I showed that if ab>0 then you cannot prove [math](a+b)^2\geq ab[/math]

Please, write down the inequality and the quantity that you add to both sides of the inequality

If ab>0 => -ab<0 .But[math] (a+b)^2 >0[/math] .How can you add these two inequalities to get : [math](a+b)^2 -ab>0[/math] and consequently [math](a+b)^2>ab[/math] ??

How can you say that without a proof??

The method that i know from high school in completing the square is the following : [math]a^2+ab+b^2=a^2+2a(\frac{b}{2})+\frac{b^2}{4}-\frac{b^2}{4}+b^2[/math] which is equal to: [math] (a+\frac{b}{2})^2 +\frac{3b^2}{4}[/math]. Which is definitely greater than or equal to zero for all values of a and b. On the other hand to get : [math] (a+b)^2\geq ab[/math] you must assume [math]a^2+ab+b^2\geq 0[/math] ,which is what you want to prove

If we treat the L.H.S as quadratic we have: [math]a = \frac{-b+\sqrt{b^2-4b^2}}{2}= \frac{-b+\sqrt{3}bi}{2}[/math] .............................or....................................................... [math]a = \frac{-b-\sqrt{b^2-4b^2}}{2}= \frac{-b-\sqrt{3}bi}{2}[/math] Which makes, a ,a complex No Merged post follows: Consecutive posts merged Yes, i completed the square and it works. But what is all this about ab??

D.H the solution x=y=z=0 is the solution to every system of equations that are equal to 0. This i suppose is well known to every one of us . Here we want solutions other than those equal to zero

I do not know how even to start with this system of equations . Do you . If yes please show me. On the other hand answers like the one given by shyvera is a clever way of saying i do not know