# triclino

Senior Members

285

1. ## sequence convergence

prove that the following sequence converges: $x_{k+2} =\frac{2}{x_{k+1} + x_{k}} ,x_{1},x_{2}>0$
2. ## system of equations

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

4. ## inequality

Yes that substitution led us to infinite solutions
5. ## system of equations

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

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

Solve the following inequality: $xy+(xy)^2 +(xy)^3>14$ needless to say i have no idea how even to start this inequality
8. ## inequality

So i take it that you cannot formally analyze your own proof??
9. ## inequality

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??
10. ## inequality

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

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

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

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

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.
15. ## inequality

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 $a^2+ab+b^2\geq 0$ in your proof)
16. ## inequality

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 $a^2+ab+b^2\geq 0$ to $(a+b)^2\geq ab$ you do the following: $a^2+ab+b^2\geq 0\Longleftrightarrow a^2+ab+b^2+ab\geq ab\Longleftrightarrow (a+b)^2\geq ab$ Is $(a+b)^2\geq ab$ a well known fact??
17. ## inequality

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?
18. ## inequality

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 $(a+b)^2\geq ab$
19. ## inequality

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

If ab>0 => -ab<0 .But$(a+b)^2 >0$ .How can you add these two inequalities to get : $(a+b)^2 -ab>0$ and consequently $(a+b)^2>ab$ ??
21. ## system of equations

How can you say that without a proof??
22. ## inequality

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

If we treat the L.H.S as quadratic we have: $a = \frac{-b+\sqrt{b^2-4b^2}}{2}= \frac{-b+\sqrt{3}bi}{2}$ .............................or....................................................... $a = \frac{-b-\sqrt{b^2-4b^2}}{2}= \frac{-b-\sqrt{3}bi}{2}$ 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??
24. ## system of equations

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
25. ## system of equations

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
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