Math Olympiad Problem | AM-GM Inequality

Math Olympiad Problem | AM-GM Inequality | an example

Рет қаралды 11,201

Күн бұрын

Do you enjoy a good mathematical challenge? In this video, we explore a fascinating problem that deals with inequality. We are given two positive numbers, a and b, and a specific equation that must be satisfied. Our task is to find the minimum value of a complex expression.
In this video, we will walk you through two attempts at solving this problem. The first attempt involves using the AM-GM inequality to derive a lower bound for the expression, but we quickly discover that the bound is not achievable. In our second attempt, we employ a clever algebraic manipulation to derive a more accurate lower bound, and we show that this bound is indeed achievable.
By the end of this video, you'll have a deeper understanding of mathematical inequalities and some clever techniques for solving complex problems. So, let's dive in and explore this intriguing problem together!

Пікірлер: 40

@ronbannon Ай бұрын

Very nice job.

@chanlyelee Ай бұрын

Thank you! Cheers!

@imgsukawijaya 4 ай бұрын

Sir, help me, If i have a, b, c then we will know: a+b+c>=3(a.b.c)^(1/3) And i try use a= 1/a, b=1/b, c=1/(a+b),, Next, we have 1/a + 1/b + 1/(a+b) >= 3[1/a . 1/b. 1/(a+b)]^(1/3) Is that wrong?

@chanlyelee 4 ай бұрын

It is not wrong, but we are unable to achieve equality. For example J≥2 does not mean that the minimum value of J is 2, as J ≥10 also implies that J≥2.

@imgsukawijaya 4 ай бұрын

@@chanlyeleethx sir

@srikanthtupurani6316 3 жыл бұрын

I struggled little bit to solve it. I was reluctant to consider 1/2a, 1/2a, 1/2b,1/2b, 1/(a+b) and apply A,M inequality. I always chose things so that the right side of the inequality is a known value. Some instincts kill us when it comes to problem solving. Most frustrating thing is we always find a new inequality which is totally different from the ones we solved before. These problems appear as simple. But they can be annoying when we go in a wrong direction.

@chanlyelee 3 жыл бұрын

Keep trying and you will be more skillful.

@ultrio325 6 ай бұрын

Haha, I feel this I try every transformation that I can think of and oftentimes nothing works, it's horribly frustrating

@ultrio325 6 ай бұрын

@@chanlyeleealso hii one of my friends was your student lol

@robertveith6383 3 ай бұрын

You wrote those fractions wrong. They need grouping symbols: 1/(2a), 1/(2b).

@hivirupalihena3102 Жыл бұрын

I arrived to a+b => 20 Then Multiplying both sides by ab ab(a+b) => 20ab 2000 => 20ab 100 => ab 10000 => (ab)^2 Taking the reciprocal on both sides 1/10000 =>1/(ab)^2 Multiplying both sides by ab(a+b) ab(a+b)/10000 => ab(a+b)/(ab)^2 2000/10000 => a+b/ab 1/5 =>1/a + 1/b Since a+b => 20 0 < 1/(a+b) 1/5 each sides gives 1/5

@chanlyelee Жыл бұрын

Great to hear from you. Here is the mistake "10000 => (ab)^2 Taking the reciprocal on both sides 1/10000 =>1/(ab)^2" Note that if x>y>0 then 0

@hivirupalihena3102 Жыл бұрын

Thank you

@Generalist18 Жыл бұрын

Easier solution. Using AMGM a+b>=20, but observe that inorder for 1/a+b to be maximum, a+b must minimum therefore a+b=20.using the given info(ab(a+b) =2000) solve for ab, given a+b=20,ab=100.add the expression 1/a +1/b +1/a+b by taking Lcm, yields {(a+b) ^2+ab}/ab(a+b),ab(a+b)=2000,ab=100,a+b=20,plug in those value, and you will get 1/4

@chanlyelee Жыл бұрын

Thanks for sharing.

@robertveith6383 3 ай бұрын

1/(a + b) needs to be written this way, for example.

@ertanyildirim6263 Жыл бұрын

Thank you so much ..very nice solutoon

@chanlyelee Жыл бұрын

Most welcome

@srikanthtupurani6316 3 жыл бұрын

I did the same thing. The greatest pain in inequalities is finding the value where maximum is attained. People underestimate these problems they think they are simple. Direct application of standard inequalities won't work. If we don't want to struggle. We can use calculus. I have seen your solution.same thought process as mine. I thought you may come up with a different solution. I feel we can solve it using convex functions.

@chanlyelee 3 жыл бұрын

checking the equality is an important step in this kind of questions.

@jokarmaths7771 2 жыл бұрын

amazing ...kzbin.info/www/bejne/npXKp6iFh9N4n6M