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Question Number 98022 by Ar Brandon last updated on 11/Jun/20
Derive the relation between an Arithmetic Mean  and a Geometric Mean  ((x_1 x_2 ...x_n ))^(1/n) ≤((x_1 +x_2 +∙∙∙+x_n )/n) ∀n∈N^∗ , ∀(x_1 ,x_2 ,...x_n )∈(R_+ ^∗ )^n
$$\mathcal{D}\mathrm{erive}\:\mathrm{the}\:\mathrm{relation}\:\mathrm{between}\:\mathrm{an}\:\mathcal{A}\mathrm{rithmetic}\:\mathcal{M}\mathrm{ean} \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathcal{G}\mathrm{eometric}\:\mathcal{M}\mathrm{ean} \\ $$$$\sqrt[{\mathrm{n}}]{\mathrm{x}_{\mathrm{1}} \mathrm{x}_{\mathrm{2}} …\mathrm{x}_{\mathrm{n}} }\leqslant\frac{\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\centerdot\centerdot\centerdot+\mathrm{x}_{\mathrm{n}} }{\mathrm{n}}\:\forall\mathrm{n}\in\mathbb{N}^{\ast} ,\:\forall\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{x}_{\mathrm{2}} ,…\mathrm{x}_{\mathrm{n}} \right)\in\left(\mathbb{R}_{+} ^{\ast} \right)^{\mathrm{n}} \\ $$
Answered by Rio Michael last updated on 11/Jun/20
The geometric mean G = (√(ab))    The arithmetic mean A = ((a+b)/2)   A−G > 0   A − G = ((a + b)/2) + (√(ab))                  = ((a + b−2(√(ab)))/2)                   = ((((√a) −(√b) )^2 )/2)  for a quadratic x^2 −2Ax + G^2  = 0 where G^2  = (√(ab)) and A = ((a +b)/2)  has roots x = A ± (√(A^2 −G^2 ))
$$\mathrm{The}\:\mathrm{geometric}\:\mathrm{mean}\:{G}\:=\:\sqrt{{ab}}\: \\ $$$$\:\mathrm{The}\:\mathrm{arithmetic}\:\mathrm{mean}\:{A}\:=\:\frac{{a}+{b}}{\mathrm{2}} \\ $$$$\:{A}−{G}\:>\:\mathrm{0}\: \\ $$$${A}\:−\:{G}\:=\:\frac{{a}\:+\:{b}}{\mathrm{2}}\:+\:\sqrt{{ab}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{a}\:+\:{b}−\mathrm{2}\sqrt{{ab}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\left(\sqrt{{a}}\:−\sqrt{{b}}\:\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{quadratic}\:{x}^{\mathrm{2}} −\mathrm{2}{Ax}\:+\:{G}^{\mathrm{2}} \:=\:\mathrm{0}\:\mathrm{where}\:{G}^{\mathrm{2}} \:=\:\sqrt{{ab}}\:\mathrm{and}\:{A}\:=\:\frac{{a}\:+{b}}{\mathrm{2}} \\ $$$$\mathrm{has}\:\mathrm{roots}\:{x}\:=\:{A}\:\pm\:\sqrt{{A}^{\mathrm{2}} −{G}^{\mathrm{2}} } \\ $$
Commented by Ar Brandon last updated on 11/Jun/20
Thanks for your time. But actually, the question requires  us to derive the above relation, instead of solving for x.
$$\mathcal{T}\mathrm{hanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}.\:\mathcal{B}\mathrm{ut}\:\mathrm{actually},\:\mathrm{the}\:\mathrm{question}\:\mathrm{requires} \\ $$$$\mathrm{us}\:\mathrm{to}\:\mathrm{derive}\:\mathrm{the}\:\mathrm{above}\:\mathrm{relation},\:\mathrm{instead}\:\mathrm{of}\:\mathrm{solving}\:\mathrm{for}\:\mathrm{x}. \\ $$

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