Prove, for abc=1 and a,b,c>0, that (a+b+c)>=3

[Tuugii]

If (a*b*c)=1 and a>0, b>0, c>0 then prove that (a+b+c)>=3.

thanks, Tuugii

 

[pka]

The following is a well known inequality:

\(\displaystyle \L\frac{{a_1 + a_2 + \cdots + a_n }}{n} \ge \sqrt[n]{{a_1 a_2 \cdots a_n }}.\)

It is very tedious to prove, done by mathematical induction.
You problem follows for the case n=3.

 

[TchrWill]

If (a*b*c)=1 and a>0, b>0, c>0 then prove that (a+b+c)>=3.

1/n x n x 1 = 1

1/n + n + 1 =>3

n^2 - 2n + 1 = 0

n = [2+/-sqrt(4 - 4)]/2 = 2/2 = 1.

1/1 + 1 + 1 = 3

Any value of n greater than 0 will result in 1/n + n + 1 =>3.

 

[pka]

If (a*b*c)=1 and a>0, b>0, c>0 then prove that (a+b+c)>=3.
1/n x n x 1 = 1
Any value of n greater than 0 will result in 1/n + n + 1 =>3.

Sorry to tell you this Will but that is not a proof.
It is a mere demonstration of one particular case and not the general case.

What about \(\displaystyle a = \frac{\pi }{e},\;b = \frac{{2e}}{{3\pi }},\quad \& \;c = \frac{3}{2}?\)

 

[Tuugii]

thanks guys!

I've spent several hours to get the following results:

1. let's consider the problem from the opposite, which is (a+b+c)<3, if a,b,c>0 and abc=1.

2. Multiply both sides of the inequality by ab : (a^2)b+a(b^2)+abc-3ab<0

3. A little algebra can convert it into this inequality : (b^2)a+b((a^2)-3a)+1<0

4. For this quadratic inequation it is clear that it has two real solutions so the discriminant D>0. Which is: (((a^2)-3a)^2)-4a>0

5. And with help of little algebra we obtain this: ((a^2)-1)^2(a-4)>0.

6. It says that a>4, in this case a+b+c<3 is impossible, thus a+b+c>=3.

Tuugii

 

[morson]

I like the way you're thinking by trying to prove it by contradiction, but your step 3 seems to be incorrect. How did you get to that?

Anyway, three positive real numbers multiply to 1, right? Which must mean they're all 1 (in which case their sum is 3) or, if they're not all 1, then at least one of them is a fraction. Can you go from there?

 

[Tuugii]

thanks morson, I'll try that way also.

algebra in step 2-3:

2. Multiply both sides of the inequality by ab : (a^2)b+a(b^2)+abc-3ab<0
2.1 a(b^2)+(a^2)b-3ab+abc<0
2.2 a(b^2)+b((a^2)-3a)+abc<0
2.3 a(b^2)+b((a^2)-3a)+1<0
3. A little algebra can convert it into this inequality : (b^2)a+b((a^2)-3a)+1<0

hope you've got it,
Tuugii

 

[morson]

Ah, I didn't know what you meant by writing "abc" as 1, but then I kicked myself and read your original post! So you've made the assumption that a quadratic in b has two solutions for any a and b (not necessarily), and solved a quartic inequality, which is making it harder than it actually is. But nonetheless, you came up with a contradiction: a is bigger than or equal to four, which contradicted your original assertion. So good job.

 

[pka]

It is really too bad that you did not try to prove directly The Arithmetric-mean-Gemetric-mean Inequality. I suspect that is the whole point of the exercise. It you have access to a reasonably good library there is a nice and elementary proof in An Introduction to Inequalities by Beckenbach and Bellman.
There is a goog proof in the above book.

 

[Tuugii]

Thanks pka,
I really want to know that way. Is that Koshi's inequality?..

Where can I find the proof? on which topic?
thanks in advance,
Tuugii