Find the Minimum… Without Expanding Everything 🤯| Nigerian Math Olympiad

Added: 2026-05-05

[00:00:02]Today we'll be solving one very very interesting question from Nigerian Mathematical Olympiad. Here we have f x given. It is one rational function.

[00:00:15]It is said to find the minimum value of f x.

[00:00:20]Given that x is positive.

[00:00:24]So we will consider numerator over denominator.

[00:00:30]Now we are going to simplify numerator first.

[00:00:34]So we will write here numerator n equal to x + 1 over x whole power 6 x power 6 + 1 over x power 6 - 2 which will be equal to x + 1 over x whole power 6 - we will take common. In another bracket we'll write x power 6 + 1 over x power 6 + 2 which will be written as x + 1 over x whole power 6 x power 6 we can write x cube whole square + 1 over x power 6 we can write 1 over x cube whole square + 2 can be written as 2 times x cube times 1 over x cube.

[00:01:33]x cube from numerator denominator we'll get over we'll still get 2.

[00:01:39]And if we consider x cube is a 1 over x cube is b then here we have a square plus b square plus 2 a b which we know equal to a plus b whole square.

[00:01:57]So our numerator will become x + 1 over x whole power 6 x cube plus 1 over x cube whole square.

[00:02:16]Now here we have power 6 x + 1 over x whole power 6. Can we factor 6 as 2 times 3?

[00:02:25]So I will write here n equal to x + 1 over x and then there is power 6 which we will write 3 times 2.

[00:02:38]x cube + 1 over x cube whole square.

[00:02:46]We know that a power b times c can be written as a power b whole power c.

[00:02:56]So we will write numerator n x + 1 over x whole cube and then we'll put whole power 2.

[00:03:12]We'll write x cube + 1 over x cube whole square.

[00:03:20]If we will consider x + 1 over x whole cube p x cube + 1 over x cube q then here we have p square - q square form.

[00:03:34]So we can utilize difference of two squares formula and we can write p + q times p - q.

[00:03:44]So we will write numerator n will be equal to x + 1 over x whole cube + x cube + 1 over x cube in one bracket times x + 1 over x whole cube x cube + 1 over x cube in other bracket from p - q.

[00:04:19]Now this is our numerator.

[00:04:21]If we will check here we have denominator which is matching with numerator's factor x + 1 over x whole cube + x cube + 1 over x cube.

[00:04:36]Let's find f x.

[00:04:39]So we will write here f x equal to numerator over denominator.

[00:04:47]So numerator is x + 1 over x whole cube + x cube + 1 over x cube in one bracket times in second bracket we'll write x + 1 over x whole cube x cube + 1 over x cube this will be in second bracket over x + 1 over x whole cube + x cube + 1 over x cube.

[00:05:31]Now we can cancel these two terms from numerator denominator.

[00:05:37]So we will write f x will be equal to x + 1 over x whole cube x cube - 1 over x cube.

[00:05:52]Now we'll be using a + b whole cube identity over here.

[00:05:57]We can write a cube which will be equal to x cube + b cube 1 over x cube + 3 a b so 3 times x times 1 over x times a + b which is x + 1 x.

[00:06:16]Then we have - x cube - 1 over x cube.

[00:06:21]Now we will cancel + x cube - x cube + 1 over x cube - 1 over x cube x x So we will write f x will be equal to 3 times x + 1 over x.

[00:06:43]This is f x.

[00:06:45]Now we have to find minimum value of f x.

[00:06:49]So if I will write two numbers x and 1 over x as per problem given is x greater than 0.

[00:07:00]So both numbers are positive numbers.

[00:07:05]Now we know that AM is greater than or equal to GM inequality.

[00:07:12]So I will find arithmetic mean of these two numbers x and 1 over x.

[00:07:18]So we have to add both of the numbers divided by count of the numbers which is 2.

[00:07:27]This should be greater than or equal to its geometric mean which will be equal to the square root of x times 1 over x product of both.

[00:07:40]Here we will cancel x from numerator denominator so this is nothing but the square root 1. We know that the square root 1 is 1.

[00:07:50]So we have x + 1 over x over 2 greater than or equal to.

[00:07:59]Now to cancel 2 from the denominator we will multiply our inequality by 2.

[00:08:07]So we'll multiply here by 2 also.

[00:08:10]Now we can cancel 2 from numerator denominator.

[00:08:15]So we can write here the minimum value of x + 1 over x is 2.

[00:08:24]Now we will write f of x first which is equal to 3 times x + 1 over x.

[00:08:35]Now minimum value of x + 1 over x is 2.

[00:08:39]So we will write f minimum x f of x minimum value will be equal to 3 times 2 which will be equal to 6. So our answer will become 6.

[00:08:56]I hope friends you will like this video.

[00:08:59]Thank you so very much for watching. Do not forget to like, share, subscribe.

[00:09:02]Bye-bye.