you’re in the running for “mathematician with best handwriting”

As a Romanian graduating looong time ago, when the specialized mathematics gazzete called (guess the name) Gazeta Matematică was the nightmare of every student from 6th grade until end of 12th grade (high school) I never thought that I will see a day when a problem from that journal will make it to the whole world.

he speaks as if he's whispering secrets as he solves this,,, thing,, which feels right. It's one of those things you'd think you aren't meant to know as a child

Your combination of logic and math is always educational. Thanks for challenging me.

As a romanian that has the olympiad next week this video is really what i need to see. Keep up the good work 🫶🏼

I love how these problems turn from 'how the hell am I supposed to solve this' to 'it's just numbers, man' with the use of some shrewd applied logic. I don't think any channel makes me want to learn more about this than this one.

Thank you. At 68 years old I learnt something new today.

There's a funny and simple technique that I came up for this question that allows to establish both upper and lower bounds.
Lemma. If a ≤ b, then √a + √b > √(a - x) + √(b + x)  for any x in (0, a].
This statement is intuitive from graph of the square root.

This allows to play with expressions like this:
√5 + √8 > √(5-1) + √(8+1) = 2 + 3
√6 + √7 > √(6-2) + √(7+2) = 2 + 3
√12 + √13 > √(12-3) + √(13+3) = 3 + 4
These are enough to establish lower bound of 26.

Simillarly
√9 + √9 > √(9-1) + √(9+1) = √8 + √10
√9 + √9 > √(9-2) + √(9+2) = √7 + √11
√9 + √9 > √(9-3) + √(9+3) = √6 + √12
√9 + √9 > √(9-4) + √(9+4) = √5 + √13
These all together give upper bound of 27.

Best math problem channel on YouTube. Interesting problems with really clearly presented solutions.

I have absolutely no idea why this was on my feed, as i am neither romanian, nor have I ever watched any type of math related content on yputube before. In fact, I haven‘t had a math lesson in almost 10 years. But I don’t know how, my dopamine addicted instagram reel brain still watched this video to the end and managed to understand everything. Thank you so much for the cool video and your enthusiasm is really contagious!

I like your .....
''Never stop LEARNING,
Those who STOP, STOP LIVING''

Since sqrt(x) is a strictly concave function, you can use Jensen's inequality to establish that sqrt((5+6+...+13)/9) = sqrt(9) = 3 > (sqrt(5) + sqrt(6) + ... + sqrt(13))/9 = S/9, i.e. 3 > S/9.
Therefore S < 27.

Olympiad: without using any operations that directly mathematically lead to the answer, compute the answer.

14:24 we are indeed, cool

Man, it's nice looking at the comment section of a video and seeing smart people.

Hello, great video! Here is my solution, it is almost geometrical, hah.
P.S. English isn't my native language, so my apologies, if some phrases sound strange and off.

I use the letter N for this sum: N = 5^0.5 + 6^0.5 + ... + 13^0.5.

Proof: N > 26

1) Let's look at graphic of function x^0.5, it's convex. Let's watch closer on interval [5; 13]. We can split it on unit-length intervals [5;6], [6;7], etc.
For every interval [n; n+1] build a right trapezoid, where one side is this interval, and two other sides - line segments from ends of the interval to according function graphics points. Every such trapezoid will have side with length 1 and two sides with lengthes n^0.5 and (n+1)^0.5. The square of it is equal to
(n^0.5 + (n+1)^0.5)/2 * 1.
If we will sum up all squares of all trapezoids, we will get value, which we call S:
S = (5^0.5 + 6^0.5)/2 + (6^0.5 + 7^0.5)/2 + ... + (12^0.5 + 13^0.5)/2.
Now we can see, that
S = N - (5^0.5)/2 - (13^0.5)/2,
or
N = S + (5^0.5 + 13^0.5)/2

2) Now, look at interval [5; 9] and build trapezoid for it the same way we did before. The square S1 can be calculated as:
S1 = (5^0.5 + 9^0.5)/2 * (9 - 5).
The same is true for [9; 13];
S2 = (9^0.5 + 13^0.5)/2 * (13 - 9).
Because of convexity of graphic of function, the square S1 will be smaller, then sum of squares of trapezoids, built on [5;6], [6;7], ..., [8; 9]. it is most noticable, if you build graphic and build all those trapezoids. The same is true for S2 and trapezoids [9; 10], [11, 12], [12, 13]. Then:
S >= S1 + S2 = 12 + 2*(5^0.5 + 13^0.5)
Now use this enequality in equality with N and S, and we will get:
N >= 12 + 2.5 *(5^0.5 + 13^0.5).

3) All is left to deal with v=(5^0.5 + 13^0.5) and give to it some estimate.
At first, let's look at its square: v^2 = 18 + 2*(65^0.5) > 18 + 2 * 8 = 34.
Then v > 34^0.5 = 34 / 34^0.5 >= 34 / 36^0.5 = 17 / 3.

Let's put this estimate:
N > 12 + 2.5 * 17 / 3 = 26 +(1/6) > 26.


Proof N < 27

Let's make 2-dimensional vector (N N) and look at it closer. We can write this vector in this way:
(N N) = ((5^0.5 + 6^0.5 + ... + 13^0.5) (13^0.5 + 12^0.5 + ... + 5^0.5)),
then we can write it as a sum of vectors:
(N N) = (5^0.5 13^0.5) + (6^0.5 12^0.5) + ... + (13^0.5 5^0.5)
Now the norm of vector (N N) is:
|(N N)| = |(5^0.5 13^0.5) + (6^0.5 12^0.5) + ... + (13^0.5 5^0.5)|
Now use the rule: the norm of sums of vectors lower or equal, then sum of those vectors norms; in our case strictly lower, because equality can be echieved, only if vectors are looking at same directions.
|(N N)| < |(5^0.5 13^0.5)| + |(6^0.5 12^0.5)| + ... + |(13^0.5 5^0.5)| = 18^0.5 + 18^0.5 + ... + 18^0.5 = 18^0.5 * 9 = 27 * 2^0.5
But also  |(N N)| = 2^0.5 * N, if we just calculate it, so:
2^0.5 * N < 27 * 2^0.5
N < 27


P.S. for N > 26 it is more understandable, if you will try to draw graphic and trapezoids it on paper ^^

Amazingly beautiful handwriting, Sir ! Congratulations ! This makes a Math teacher even better !

This man has an extremely rare combination of talents - audience connection, charisma, vocal control, beautiful writing and - of course - an exceptional ability to explain complex mathematical ideas as clearly and understandably as possible. And all of this done with a quiet and confident dignity and without overt showmanship!

In the first part, each pair around √9 adds up to less than 6, proving that the total <27. After this I tried pairing them differently to show that pairs are > an integer. It's easy to show using a similar technique that √5 + √8 >5, and √6 + √7 >5. The others do not pair up so neatly, so I ended by multiplying out (√10 +√11 +√12 +√13)^2; this turns out to be greater than 176, which is greater than 13^2. Therefore √10 +√11 +√12 +√13 > 13. So the whole sum of the nine roots is greater than 5 + 5 + 3 + 13, i.e. greater than 26, QED. Thanks for teaching a new technique!

The really scary thing is that problem used to be for the 8th grade. Not any high school intricacies. Just plain 8th grade algebra and inequalities. That high the level of math it used to be some 30 - 40 years ago in Romania. Compared with the sheer mediocrity nowadays, it's scary. Thanks for the elegant solution. Greetings from Bucharest, from a mathematics Olympian in the 80s.