practice problems pending

 Q1.

S1.
First by trial and error.
2^(1/2) > 1.
2^(1/2) < 3^(1/3). Why?
Raise both to powers 6.
2^3 < 3^2 => 8 < 9 correct.

So 3^(1/3) is biggest so far.
Now 3^(1/3) > 4^(1/4) why?
Raise both to powers 12.

3^4 > 4^3 => 81 > 64. True.

So looks like it could be 3^(1/3) since it's bigger than both its neighbors.

Second by calculus.
How to differentiate x^(1/x) and find its critical points?

y = x^(1/x) =>  ln(y) = f(y) = (1/x).ln(x)

To differentiate f(y) w.r.t. 'x' use chain rule.

d(f(y))/dx = d(f(y))/dy . dy/dx

d(f(y))/dy = 1/y

=> LHS = dy/dx.1/y

Now differentiate RHS wrt 'x'.
Using quotient rule.

What is quotient rule?
f(x) = u(x)/v(x) => f'(x) = [v(x).u'(x) - u(x).v'(x)]/(v(x))^2

So f'(RHS) = [x.1/x - ln(x).1]/x^2
for it to be 0:
1 - ln(x) = 0 => x = e

So critical point is at 'e'.
At x < e slope > 0
At x > e slope < 0

=> There is a maxima at 'e'.
But x has to be an integer.

So compare values at 2 and 3.
Why?
Since 2 < e < 3

Which gives us the answer as 3.


Third using algebra.
Let us check when the next term becomes smaller than the previous term.

(n+1)^(1/n+1) < n^(1/n)
Raise powers:

(n+1)^n < n^(n+1)
=>
(n+1/n)^n < n
=>
(1 + 1/n)^n < n

Look at LHS, its maximum value is e as n-> infinity.
So LHS max value is e.
At n = 3 it will break.


Q2.

S2.
Approach 1.
By using RMS-AM and AM-GM inequalities.
If we add all of them we get:
2.sigma(x_i) = sigma(x_i^2)

These terms remind us of RMS inequality.
It says that:
sqrt(sigma(x_i^2)/5) >= sigma(x_i)/5

Using both:
sqrt(2.sigma(x_i)/5 >=  sigma(x_i)/5

Let y = sigma(x_i)
=>
sqrt(2y/5) >= y/5 => 2y/5 >= y^2/25
=> y^2 <= 10y
y > 0 => y <= 10

So sum of all x_i <= 10, if we keep each of them as 2, all the original equations get satisfied. So this is one solution.
We have found upper bound now let's try to find lower bound.

Multiply all given equations:

(x1.x2..x5)^2 = (x1+x2).(x2+x3)...(x5+x1)

Apply AM.GM on each pair =>
(x1+x2) >= 2.sqrt(x1.x2)
Multiply all =>
(x1+x2).(x+x3)...(x5+x1) >= 32.(x1x2...x5)
Substitute LHS
=> (x1.x2..x5)^2 >= 32.(x1x2...x5)
=> (x1.x2..x5) >= 32

Again AM.GM =>
(x1 + x2 ... x5)/5 >=  (x1.x2..x5)^1/5 >= 2
=> Sigma(x_i) >= 10

Overall
10 <= sigma(x_i) <= 10 => sigma(x_i) = 10

Now let's go back to our RMS AM inequality where we got y<= 10.
Since we have proved that y = 10 => RMS AM inequality had achieved equality.
That's possible only when x1 = x2 ... x5
=> each of them is 2.

Approach 2.

Assume that min(x1,x2...x5) = a and max = b
So
a^2 = x_i + x_{i+1} for some i => a^2 >= 2a => a >= 2 since a > 0
b^2 = x_j + x_{j+1} for some j => b^2 <= 2b => b <= 2 since b > 0

2 <= a <= b <= 2 => a = b = 2 => each x_i is exactly 2.