Let x ∈ R show that f ( x ) = | sin ⁡ x | + | sin ⁡ ( x + 1 ) | + | sin ⁡ ( x + 2 ) | &gt; 8 5 since f ( x ) = f ( x + π ) ,it sufficient to show x ∈ ( 0 , π ]

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Daniel Valdez · Accepted Answer

Since   sin is a concave function on   [  0  ,  π  ] and sum of concave functions is a concave function,we have      min          [      0      ,      π      ]        f  =  min  {  f  (  0  )  ,  f  (  π  −  1  )  ,  f  (  π  −  2  )  ,  f  (  π  )  }  =  f  (  π  −  1  )  =  2  sin  ⁡  1  &amp;gt;      8    5

Villaretq0 · Accepted Answer

Drafting behind first clever answer, appealing to the concavity of   sin on   [  0  ,  π  ] quickly reduces the problem to showing the inequality  2  sin  ⁡  1  &amp;gt;      8    5    .From   π  &amp;lt;      22    7   we deduce             3      π        10    &amp;lt;      66    70    &amp;lt;  1, and so  2  sin  ⁡  1  &amp;gt;  2  sin  ⁡            3      π        10    =  2  ⋅      1    4    (  1  +      5    )  &amp;gt;      8    5    .The last inequality (which itself is a reasonably tight bound on the Golden Ratio   ϕ) follows from rearranging and squaring.

Let x ∈ R show that f ( x ) = <mrow class="MJX-TeX

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