Olympiad related problem

[jaunty sky]

Question 2. (I don't have much experience in inequalities. )

[frigid knoll]

<@&1227988399579730072>

Note for OP

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[regal dove]

Did you try a reverse image search on this one?

[peak ingot]

50?

[jaunty sky]

50 is correct.

Can you share your solution?

[peak ingot]

Multiply Denominator on both sides
After multiplying
Shift 100a1 + 100a2 + 100a3 part to LHS

Now your eq looks like
a1^2 - 100a1 + a2^2 - 100a2.........+ a100^2 - 100a100 = 0
Now if you observe we can write it as perfect squares

(a1 - 50)^2 + ( a2 - 50)^2.....+ (a100 - 50)^2 -100•(50²)= 0
Now
(a1 -50)^2 + (a2 - 50)^2 +..... + (a100 - 50)^2 = 100•(50²)
now
a1 will be maximum when all other squared terms are zero
And since it isn't said that all integers are same or distinct so
a2 to a100 is 50 so (50 - 50)² becomes 0
So leaving us with
(a1 - 50)^2 = 100•(50²)
Sq root both sides
a1 = 10•50 + 50
a1 = 550
550/11 = 50

[jaunty sky]

Oh nice.

Thanks a lot @peak ingot .

+solved @peak ingot