Is (a+b+c)2>0 the correct approach for proving a2+b2+c2>ab+bc+ca?

[azizlwl]

Homework Statement

a. Prove: If a≠b≠c are real numbers, then a²+b²+c²>ab+bc+ca
b. Prove: If a>0, b>0and a≠b, then a/b+b/a>2

Homework Equations

(real numbers)²>0

The Attempt at a Solution

a.
(a+b+c)²>0
a²+b²+c²>-2(ab+bc+ca)
Try to prove -2(ab+bc+ca) > ab+bc+ca but not true, -2.4 <4


Must be wrong approach?

b.
Same approach with same result.
a/b+b/a>0
Since a>0, b>0 =>ab>0
(a+b)²/(ab)>2

Freeze...no idea

 

[mtayab1994]

Mod note: Removed quoted work.[/color]
Hello, Ahmed you're not supposed to give people the answer to the problem. You have to 'help them solve'. And by the way the first one you can do a proof by contradiction.

 

[AliAhmed]

Oh, I'm so sorry, I am new to the forum and have only skimmed over the rules. Thank you for notifying me, I will read the entire forum rules so I don't make such mistakes again.

 

[mtayab1994]

Oh, I'm so sorry, I am new to the forum and have only skimmed over the rules. Thank you for notifying me, I will read the entire forum rules so I don't make such mistakes again.

Ok not a problem.

 

[azizlwl]

Thank you Mr. Ali Ahmed.
I've racking my brain(old man's brain) to max. to find solution.

 

[azizlwl]

Lessons learned from Mr Ali.

1.
(a±b)>0 wrong
instead
(a±b)≥0

2. Many ways of getting a², ab, b²...
My only way (a+b+c)²
but there are other ways
(a-b)², (b-c)²,(c-a)²...

from lessons learn, now i can solve 3rd problems

3. Prove: If a² + b²=1 and c² + d²=1,
then, ac + bd ≤ 1

Way of getting ac, (a-c)² ≥ 0
bd, (b-d)² ≥ 0

a²+c²-2ac ≥0
b²+d²-2bd ≥0

a²+c² +b²+d² -2(ac+bd)≥0

-2(ac+bd)≥-2
Changing sense of equalityThus, ac+bd ≤ 1

 

[AliAhmed]

Exactly azizlwl.

The main problem in math proofs is that you sort of have to trick the solution into appearing (which is usually not obvious). But maybe for future reference, if you ever encounter such problems again, the way I start is by coming up with ways of getting what I want (necessary terms, number, etc.) in the equation and then reduce the equation to its simplest form.

What might often happen is that the reduction turns out to be useless, so when I reach such a point in my derivation I disregard my last method of obtaining my necessary terms, and use a new one (if you are confident that you made no mistake in the reduction and you arrive at a useless conclusion, then start fresh).

But I understand where your coming from, I often go crazy even trying to solve the simplest problems.

 

[azizlwl]

Here difference way done by Dr. Mark.
a/b+b/a>2
Since both a and b are positive, multiply through by ab
a^2+b^2>0
a^2-2ab+b^2>0
(a-b)^2>

Then i asked him what is the difference and which is easier.
Here his answer


"Typically you start with what is given and derive a truth, but if you can start with the truth and derive what is given, then that is valid as well. Usually it is easier the first way, because you may not know what to begin with the second way."