\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
First, we notice that \[ G_{k + 1}^{k + 1} = \prod _{t = 1}^{k + 1} a_t = a_{k + 1} G_{k}^{k}, \] and hence \[ G_{k + 1} = \left (a_{k + 1} G_{k}^{k}\right )^{\frac {1}{k + 1}}. \]
Similarly, notice that \[ (k + 1) A_{k + 1} = \sum _{t = 1}^{k + 1} a_t = a_{k + 1} + k A_{k}. \]
Hence, \begin {align*} (k + 1)\left (A_{k + 1} - G_{k + 1}\right ) & \geq k\left (A_k - G_k\right ), \\ a_{k + 1} + k A_k - (k + 1) \left (a_{k + 1} G_{k}^{k}\right )^{\frac {1}{k + 1}} & \geq k a_k - k G_k, \\ a_{k + 1} + k G_k & \geq (k + 1)a_{k + 1}^{\frac {1}{k + 1}} G_{k}^{\frac {k}{k + 1}}. \end {align*}
Dividing both sides by \(G_k\), we have \begin {align*} \frac {a_{k + 1}}{G_k} + k & \geq (k + 1) a_{k + 1}^{\frac {1}{k + 1}} G_k^{- \frac {1}{k + 1}}, \\ \lambda _k^{k + 1} + k & \geq (k + 1) \left (\frac {a_{k + 1}}{G_k}\right )^{\frac {1}{k + 1}}, \\ \lambda _k^{k + 1} + k & \geq (k + 1) \lambda _k, \\ \lambda _k^{k + 1} - (k + 1) \lambda _k + k & \geq 0, \end {align*}
as desired. (Notice that the condition for the equal sign is equivalent as well.)
By differentiation, we have \[ f'(x) = (k + 1)x^k - (k + 1) = (k + 1) (x^k - 1). \]
When \(x \in (0, 1), x^k \in (0, 1), f'(x) < 0\), and hence \(f\) is strictly decreasing.
When \(x \in (1, \infty ), x^k \in (1, \infty ), f'(x) > 0\), and hence \(f\) is strictly increasing.
Hence, \(f(1)\) is the minimum for \(f\) on \((0, \infty )\). This means for all \(x \in (0, \infty )\), we have \[ f(x) \geq f(1) = 1^{k + 1} - (k + 1) + k = 0, \] taking the equal sign if and only if \(x = 1\).
We show this by induction. For the base case \(n = 1\), \(A_1 = G_1 = a_1\), so naturally \(A_n \geq G_n\) is satisfied.
Assume that the statement holds for some \(n = k\), i.e. \(A_k \geq G_k\), \(A_k - G_k \geq 0\). Since \(k > 0\) as well, we must have \[ (k + 1)(A_{k + 1} - G_{k + 1}) \geq k (A_k - G_k) \geq 0. \]
We also have \(k + 1 > 0\), and hence \[ A_{k + 1} - G_{k + 1} \geq 0 \iff A_{k + 1} \geq G_{k + 1}, \] meaning the statement holds for \(n = k + 1\) as well.
Hence, by the principle of mathematical induction, we must have \(A_n \geq G_n\) for all \(n \in \NN \), which finishes our proof.
We show this by induction. For the base case \(n = 1\), this condition is naturally satisfied.
Assume that the statement holds for some \(n = k\), i.e. \(A_k = G_k \implies a_1 = a_2 = \cdots = a_k\). We show this for \(n = k + 1\). If \(A_{k + 1} = G_{k + 1}\), then we must have \[ k (A_k - G_k) \leq (k + 1)(A_{k + 1} - G_{k + 1}) = 0, \] but since \(A_k \geq G_k\), we must have then \(A_k = G_k\), and hence the equal sign in the inequality being taken.
This must mean that \[ \lambda _{k} = \left (\frac {a_{k + 1}}{G_k}\right )^{\frac {1}{k + 1}} = 1, \] and hence \[ a_{k + 1} = G_k. \]
At the same time, since \(A_k = G_k\), we must have \(a_1 = a_2 = \cdots = a_k\), and hence \(G_k = a_1 = a_2 = \cdots = a_k\). Therefore, we must also have \[ a_1 = a_2 = \cdots = a_k = a_{k + 1}, \] which proves the statement that \(A_{k + 1} = G_{k + 1}\) implies \(a_1 = a_2 = \cdots = a_k = a_{k + 1}\), which is the original statement for \(n = k + 1\).
Hence, by the principle of mathematical induction, we must have \(A_n = G_n\) implies \(a_1 = a_2 = \cdots = a_n\) for all \(n \in \NN \), which finishes our proof.