Math Olympiad Problem | AM-GM Inequality | an example

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!

[srikanthtupurani]

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.

[ertanyildirim]

Thank you so much ..very nice solutoon

[hivirupalihena]

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) <= 1/20

Adding the inequality 1/a + 1/b => 1/5 each sides gives

1/5<= 1/a + 1/b + 1/(a+b)
Where did I gone wrong

[srikanthtupurani]

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.

[Generalist]

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

[imgsukawijaya]

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?

[ignacioaraya]

Why 1/a has to be equal to 1/b and 1/(a+b) ? 😢

[Watermelon-jtpc]

first am gm inequality that i can solve