Neat. The 4th method (i.e. the one I used) is (a+b)(b+c)(a+c)/(abc) = a/c + b/c + a/b + ... = 2 + a(1/b + 1/c) + symmetric = 2 + [a (1/a + 1/b + 1/c) -1 + symmetric] = -1 + (a+b+c)(1/a + 1/b + 1/c) then use AM >= HM, which in this case gives you LHS > = -1 + 9 =8
(x+y)/2>=sqrt(xy), so a+b>=2sqrt(ab), b+c>=2sqrt(b+c) and c+a>=2sqrt(ca). Multiply all three and simplify the right side: (a+b)(b+c)(c+a)>=2sqrt(ab)2sqrt(bc)2sqrt(ca)=8sqrt(aabbcc)=8abc, then divide both sides by abc. BOOM. Done.
During our days A.M.>=G.M. was in high school's syllabus, and this question appeared in past public exam for the topic. The 1st method used by the host proves the famous inequality first and then applies it 3 times.
(a+b)(b+c)(c+au)/abc = ((a+b)/a)((b+c)/b)((c+a)/c) = (1+b/a)(1+c/b)(1+a/c) Assuming c>=b>=a It turns out that each of these factors is greater or equal to 2 So its product must be greater or equal to 8. Is that a 4th method? Thank you.
My Idea would be to calculate the gradient of the expression as a three dimensional function. Then show that at the point of the minimum value of the function, i.e. where the gradient ist a zero vector, ist 8.
My first method is to prove that a+b, b+c and c+a are greater or equal to 2abc. (it's just an idea that I haven't do 😁). Thanks for your first method which show me how to go about this. And my second method is your 3rd method 😅. Too long and painful 😂
We know that AM ≥ GM so Let a b c be the terms in the certain sequence of positive integers so a+b/2 ≥√ab Similarly b+c/2≥√bc ; a+c/2≥√ac We can multiple those above inequalities We get (a+b)(b+c)(a+c)/8≥√a²b²c² Then (a+b)(b+c)(a+c)/8≥abc Now (a+b)(b+c)(a+c)/abc≥8 when a,b,c >0 Hence proved 😜
You wrote your post wrong. Whenever you wrote radical signs with more than one variable, such as ab, you need grouping symbols around the ab. Also, whenever you divide by a product, such as abc, you must put it inside grouping symbols.
@@fahrenheit2101 -- Wrong! Go back and delete your post, since you completely do not understand what it means to write correct notation. You don't get credit for what is "meant." What you replied is stupid. And, if a similar corrective post shows up again for a user, *you* stay out of it as you do not know what you are talking about. Not being a formal paper is irrelevant. Get that through your head sooner than later.
@@robertveith6383 How about... no? You don't get to tell me what I do and don't post. It's also generally nicer to actually explain yourself when correcting someone, rather than going "NO! YOU ARE WRONGGGGG! DELETE IT!!!".
@@robertveith6383 Lmao just went back and checked what I said. You are... either seriously bored and sarcastic, or seriously troubled. Hoping it's the former. If it's the latter... wow.
Neat. The 4th method (i.e. the one I used) is (a+b)(b+c)(a+c)/(abc) = a/c + b/c + a/b + ... = 2 + a(1/b + 1/c) + symmetric = 2 + [a (1/a + 1/b + 1/c) -1 + symmetric] = -1 + (a+b+c)(1/a + 1/b + 1/c) then use AM >= HM, which in this case gives you LHS > = -1 + 9 =8
using AG on a,b ; a,c ; b,c and multiply it all
(x+y)/2>=sqrt(xy), so a+b>=2sqrt(ab), b+c>=2sqrt(b+c) and c+a>=2sqrt(ca). Multiply all three and simplify the right side: (a+b)(b+c)(c+a)>=2sqrt(ab)2sqrt(bc)2sqrt(ca)=8sqrt(aabbcc)=8abc, then divide both sides by abc. BOOM. Done.
I did it the same way!
Cosi: (a+b)(b+c)(c+a) >= 2sqrt(ab)2sqrt(bc)2sqrt(ca) = 8abc.
Just multiply all the brackets of numerator and apply AM >= GM on those 8 numbers
He won't do that, it's way too simple, he loves complicating things to show how much math he knows.
During our days A.M.>=G.M. was in high school's syllabus, and this question appeared in past public exam for the topic. The 1st method used by the host proves the famous inequality first and then applies it 3 times.
A fourth way:
We want to prove that (a+b)(b+c)(c+a)
At a glance I'd say AM ≥ GM might help here
Sir solve this problem find the equation of the circle inscribed in triangle whose sides are the lines L1:x+y=8 L2:2x+y=22 L3: 3x+y=22
(a+b)(b+c)(c+au)/abc =
((a+b)/a)((b+c)/b)((c+a)/c) =
(1+b/a)(1+c/b)(1+a/c)
Assuming c>=b>=a
It turns out that each of these factors is greater or equal to 2
So its product must be greater or equal to 8.
Is that a 4th method?
Thank you.
Awesome and beautiful solutions loved them three
thanks so much for the daily videos keep it up freind you rock!
You are The Premier Syber Man!
Thank you!!! 🥰
My Idea would be to calculate the gradient of the expression as a three dimensional function. Then show that at the point of the minimum value of the function, i.e. where the gradient ist a zero vector, ist 8.
Nice solutions.thanks.
Beautiful
Can you do this via induction?
My first method is to prove that a+b, b+c and c+a are greater or equal to 2abc. (it's just an idea that I haven't do 😁). Thanks for your first method which show me how to go about this. And my second method is your 3rd method 😅. Too long and painful 😂
Thanks nice video
Np. Thank you!
Yo I saw this on another but it was the opposite. You were given that they're positive and had to find the maximum
What? I didn't understand the last method
ALL EQUALS 1
We know that AM ≥ GM so
Let a b c be the terms in the certain sequence of positive integers so
a+b/2 ≥√ab
Similarly
b+c/2≥√bc ; a+c/2≥√ac
We can multiple those above inequalities
We get (a+b)(b+c)(a+c)/8≥√a²b²c²
Then
(a+b)(b+c)(a+c)/8≥abc
Now
(a+b)(b+c)(a+c)/abc≥8
when a,b,c >0
Hence proved 😜
You wrote your post wrong. Whenever you wrote radical signs with more than one variable, such as ab, you need grouping symbols around the ab. Also, whenever you divide by a product, such as abc, you must put it inside grouping symbols.
@@robertveith6383 Not if it's obvious what is meant. It's a youtube comment, not a formal paper.
@@fahrenheit2101 -- Wrong! Go back and delete your post, since you completely do not understand what it means to write correct notation. You don't get credit for what is "meant." What you replied is stupid. And, if a similar corrective post shows up again for a user, *you* stay out of it as you do not know what you are talking about. Not being a formal paper is irrelevant. Get that through your head sooner than later.
@@robertveith6383 How about... no? You don't get to tell me what I do and don't post. It's also generally nicer to actually explain yourself when correcting someone, rather than going "NO! YOU ARE WRONGGGGG! DELETE IT!!!".
@@robertveith6383 Lmao just went back and checked what I said. You are... either seriously bored and sarcastic, or seriously troubled. Hoping it's the former. If it's the latter... wow.
i lv u 99.9%
lv u 2 😂
Again, you haven't show the cruelty of those inequality problems yet. This's too easy.
Maybe you'll like this one better?
for all positive integers b > a, show that a↑b > b↑a,
where the ↑ is the exponential operator.
@ OP - This is "too easy!?" You wrote that *after* this video *and* without showing your own
solution here. Your words are hollow.
@@GirishManjunathMusic not true
@@duongquocthongho2117 I'd say prove it but that was my fault I didn't state the full statement. It does work for positive integers greater than 1.
@@GirishManjunathMusic a=2;b=3