Daily questions _190908

Known number of columns $ {a_n} $ satisfy: A_1 = $ \ dfrac. 1 {{}} $. 3, \ (n-A_ {+}. 1 = (. 1-A_N) \ SiN A_N \) , n-$ \ in \ mathbb { . N} ^ \ ast $ Prove:
$ (. 1) $ $ 0 <A_N \ leqslant \ dfrac {. 1} {n-+ 2}, n-\ in \ mathbb {N} ^ \ AST; $
$ (2) $ $ \ cos \ sqrt {2a_1} \ cos \ sqrt {2a_2} \ cdots \ cos \ sqrt {2a_n}> \ dfrac {2} {n + 2}, n \ in \ mathbb {N} ^ \ ast. $
parsing
$ (1) $ easy to know \ [\ forall x \ in \
left (0, \ dfrac {1} {3} \ right], 0 <(1-x) \ sin x <x. \] first consider proof \ ( \ FORALL n-\ in \ mathbb {N} ^ \ AST, 0 <A_N \ leqslant \ dfrac. 1 {{}}. 3 \) .
when the foundation is clearly summarized $ n = 1 $, the above statement is true.
induction recursive assumed that when $ n = k (k \ geqslant2 \ text { and} k \ in \ mathbb {N } ^ \ ast) $ time, $ 0 <a_k \ leqslant \ dfrac {1} {3}. $ it \ [a_ {k + 1} = (1-a_k) \ sin {a_k} \ in \ left (0, \ dfrac {1} {3} \ right], \] Thus\ (\ forall n \ in \ mathbb {N} ^ \ ast, a_n \ in \ left (0, \ dfrac {1} {3} \ right]. \) followed prove $ \ forall n \ in \ mathbb { N} ^ \ ast, 0 < a_n \ leqslant \ dfrac {1} {n + 2}. $
summarized apparent foundation $ $ n = 1, the above statement is true.
induction recursive assumed that when $ n = k (k \ geqslant2 \ text {and} k \ in \ mathbb {N } ^ \ ast) $ time, \ (0 <a_k \ leqslant \ dfrac {. 1} {K + 2}. \) then \ [0 <a_ {k + 1 } = (1-a_k) \ sin {a_k} <(1-a_k) a_k \ leqslant \ dfrac {k + 1} {(k + 2) ^ 2} <\ dfrac {1} {k + 3}. \ ]
so \ (\ FORALL n-\ in \ mathbb {N} ^ \ AST, 0 <A_N \ leqslant \ dfrac {. 1} {n-+ 2} \) .
$ (2) $ original inequality can be written \ [\ COS \ sqrt {2a_1} \ cos \ sqrt {2a_2} \ cdots \ cos \ sqrt {2a_n}> \ dfrac {2} {3} \ cdot \ dfrac {3} {4} \ cdots \ dfrac {n + 1} {n +2} \]
long syndrome \ [\ forall n \ in \
mathbb {N} ^ \ ast, \ cos {\ sqrt {2a_n}}> \ dfrac {n + 1} {n + 2}. \] binding \ ((1)\)The conclusion can be seen only proved \ [\ forall n \ in \ mathbb {N} ^ \ ast, \ cos {\ sqrt {\ dfrac {2} {n + 2}}}> \ dfrac {n + 1} {n +2}. \]
order $ x = \ sqrt {\ dfrac {2} {n + 2}} \ in \ left (0, \ dfrac {\ sqrt 6} {3} \ right], $ like the formula It is equivalent to \ [\ forall x \ in \
left (0, \ dfrac {\ sqrt {6}} {3} \ right], \ cos x> 1- \ dfrac {1} {2} x ^ 2. \] It is easy to prove the above inequality. inequality in question thus proved.