Prove that sin ⁡ x + 2 x ≥ 3 x ( x + 1 ) π for all x ∈ [ 0 , π 2 ]How did π come in the expression? This is how I tried to solve. sin ⁡ x ≤ x so sin ⁡ 2 x ≤ 2 x sin ⁡ x + 2 x ≤ 3 xApplying AM GM inequality: sin ⁡ x + 2 x ≥ 2 2 x sin ⁡ x 3 x ≥ 2 2 x sin ⁡ x This how far I can go.

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Prove that   sin  ⁡  x  +  2  x  ≥            3      x      (      x      +      1      )        π   for all   x  ∈  [  0  ,      π    2    ]How did π come in the expression? This is how I tried to solve.  sin  ⁡  x  ≤  x so   sin  ⁡  2  x  ≤  2  x  sin  ⁡  x  +  2  x  ≤  3  xApplying AM GM inequality:   sin  ⁡  x  +  2  x  ≥  2      2    x    sin    ⁡    x    3  x  ≥  2      2    x    sin    ⁡    x  This how far I can go.

Mya Hurst · Accepted Answer

sin    ⁡    x    +    2    x    ≥          (      2      +              2        π            )        x    for       x    ∈          [      0      ,              π        2            ]       , Since   sin  ⁡  x  +  2  x is a concave functionTo prove             (      2      +              2        π            )        x    ≥                  3        x        (        x        +        1        )            π       in the range, just verify that                     3        x        (        x        +        1        )            π       is a convex functionand the inequality holds at end points.

Alani Conner · Accepted Answer

Just expanding first answer, over   I  =      (    0    ,          π      2        )   we have:                    (1)                                          d            2                                d                          x              2                                                (          sin          ⁡          x          +          2          x          )                =        −        sin        ⁡        x        &amp;lt;        0            Hence   f  (  x  )  =  sin  ⁡  (  x  )  +  2  x is a concave function, and since:                    (2)                                          d            2                                d                          x              2                                                (                      3            π                    x          (          x          +          1          )          )                =                  6          π                &amp;gt;        0              g  (  x  )  =      3    π    x  (  x  +  1  ) is a convex function. We have   f  (  0  )  =  g  (  0  )  =  0 and since   π  &amp;gt;  2:                    (3)                    f                  (                      π            2                    )                =        π        +        1        &amp;gt;                  3          2                +                              3            π                    4                =        g                  (                      π            2                    )                    proving that:                    (4)                    ∀        x        ∈        I        ,                f        (        x        )        &amp;gt;                  (          2          +                      2            π                    )                x        &amp;gt;                  (                      3            2                    +                      3            π                    )                x        &amp;gt;        g        (        x        )        .

Prove that sin ⁡ x + 2 x ≥

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