Two inequalities Let$n$ be an integer, $n\geq 2$ and $a_1,a_2,...,a_n$ positive real numbers.
For each$\{2,3,...,n\}$ we define the numbers as follows:
$s_k=a_1a_2...a_k)^{1/k}$.
a.) Pro thafor all$k$ in the set $\{2,3,...,n-1\}$ we have $s_{k+1}\geq s_k$
b.) Prove, that $$eq i<j\leq n} (\sqrt{a_i}-\sqrt{a_j})^2$$ (CALCULUS allowed)
 A: a) $s_{k+1} - s_k = a_{k+1} - (k+1)(a_1a_2...a_{k+1})^{\frac{1}{k+1}} + k(a_1a_2...a_k)^{\frac{1}{k}}$.
Apply AM-GM inequality we have: $a_{k+1} + k(a_1a_2...a_k)^{\frac{1}{k}} \geq (k+1)(((a_1a_2...a_k)^{\frac{1}{k}})^ka_{k+1})^{\frac{1}{k+1}} = (k+1)(a_1a_2...a_{k+1})^{\frac{1}{k+1}}$, this means: $s_{k+1} \geq s_k$.
b) Assume without loss of generality that $0 < a_1 \leq a_2 \leq ...\leq a_n$, then:
$\displaystyle \max_{1\leq i < j\leq n} \left(\sqrt{a_i} - \sqrt{a_j}\right)^2 = \left(\sqrt{a_n} - \sqrt{a_1}\right)^2 = a_1 - 2\sqrt{a_1a_n} + a_n$. Thus we need to prove:
$a_2 + a_3 + a_4 + ... + a_{n-1} - n(a_1a_2...a_n)^{\frac{1}{n}} + 2\sqrt{a_1a_n} \geq 0  $
Consider the function $f(a_2, a_3,...,a_{n-1}) = a_2 + a_3 +...+ a_{n-1} - n(a_1a_2...a_n)^{\frac{1}{n}} + 2\sqrt{a_1a_n}$ on the interval $[a_1, a_n]$. We now find the critical points of $f$. Setting all partial derivatives of $f$ w.r.t $a_2$, $a_3$,...,$a_{n-1}$ to $0$ we have:
$\dfrac{\partial f}{\partial a_2} = 0 \iff a_2^{n-1} = a_1a_3...a_n$
$\dfrac{\partial f}{\partial a_3} = 0 \iff a_3^{n-1} = a_1a_2a_4...a_n$
....
$\dfrac{\partial f}{\partial a_{n-1}} = 0 \iff a_{n-1}^{n-1} = a_1a_2...a_{n-2}a_n$.
So: $a_2^n = a_3^n = ...= a_{n-1}^n = a_1a_2...a_n$.
Thus: $a_2 = a_3 = ... = a_{n-1} = x$. Solve for $x$ and get:
$x^{n-1} = a_1\cdot x^{n-3}\cdot a_n \to x = \sqrt{a_1a_n}$.
Thus the critical point of $f$ is: $(a_2,a_3,...,a_{n-1}) = (x,x,...x)$.
Thus: $f_{min} = f(x,x,...x) = n\sqrt{a_1a_n} - n(a_1a_n(a_1a_n)^{\frac{n-2}{2}})^{\frac{1}{n}} = 0$. 
