Proving a Nice Inequality in Three Ways

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PLAYLISTS 🎵 :

[stmmniko]

using AG on a, b ; a, c ; b, c and multiply it all

[adandap]

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

[Lequang]

Cosi: (a+b)(b+c)(c+a) >= 2sqrt(ab)2sqrt(bc)2sqrt(ca) = 8abc.

[HoSza]

(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.

[RR-xbnh]

Just multiply all the brackets of numerator and apply AM >= GM on those 8 numbers

[hwu]

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.

[Blaqjaqshellaq]

A fourth way:
We want to prove that (a+b)(b+c)(c+a)<=8abc, or in other words that a^2*b + a^2*c + b^2*a + b^2*c + c^2 *a + c^2*b<=6abc.
(a^2*b+a^2*c)/abc=a/c + a/b, and if we do the same for the other four, all six terms divided by abc will equal a/b + a/c + b/a + b/c + c/a + c/b, so it's a matter of proving that this sum <=6. Consider f(x)=x + 1/x: f'(x)=1 -1/x^2, so f'(x) is negative when x is between 0 and 1, zero when x=1, and positive when x>1. Which means that when x is positive, f(x) is at its minimum when x=1, at which point f(x)=2. Which means that f(x)<=2 for all positive values of x. Which means that a/b + b/a<=2 when both a and b are positive. And the same is true for a/c + c/a and b/c + c/b. If we add all three, it follows that a/b + a/c + b/a + b/c + c/a + c/b<=6

[kinshuksinghania]

At a glance I'd say AM ≥ GM might help here

[mahdiali]

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

[jmart]

(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.

[SuperYoonHo]

Awesome and beautiful solutions loved them three
thanks so much for the daily videos keep it up freind you rock!

[tr]

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.

[damiennortier]

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 😂

[Johnny-twpr]

Yo I saw this on another but it was the opposite. You were given that they're positive and had to find the maximum

[fertfert]

What? I didn't understand the last method

[modern_genghis_khan]

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 😜

[angha]

Again, you haven't show the cruelty of those inequality problems yet. This's too easy.