Prove 1 2 + 1 2 ( u + 1 ) 2 − 1 1 + 2 u ≥ 0 for u ≥ 0This inequality provides a tight lower bound to 1 + 2 u for u ≥ 0 without a radical. I was trying to solve it by squaring the radical and cross-multiplying and repeated differentiation of the resulting expression, I wonder if there is a quicker solution. Thanks.

Question

Prove       1    2    +      1          2      (      u      +      1              )        2              −      1          1      +      2      u        ≥  0 for    u  ≥  0This inequality provides a tight lower bound to       1    +    2    u   for   u  ≥  0 without a radical. I was trying to solve it by squaring the radical and cross-multiplying and repeated differentiation of the resulting expression, I wonder if there is a quicker solution. Thanks.

wlanauee · Accepted Answer

It&#039;s      2    u    +    1    (      u    2    +  2  u  +  2  )  ≥  2  (  u  +  1      )    2  or  (  2  u  +  1  )  (      u    2    +  2  u  +  2      )    2    ≥  4  (  u  +  1      )    4  or      u    3    (  2      u    2    +  5  u  +  4  )  ≥  0.Done!Also, we can use the following way.Let       2    u    +    1    =  xThus,   x  ≥  1 and we need to prove that      1    2    +      1          2                        (                                                    x                2                            −              1                        2                    +          1          )                2              ≥      1    x  or  1  +      4          (              x        2            +      1              )        2              ≥      2    x  or  x  (      x    2    +  1      )    2    +  4  x  ≥  2  (      x    2    +  1      )    2  or      x    5    +  2      x    3    +  x  +  4  x  ≥  2      x    4    +  4      x    2    +  2or      x    5    −  2      x    4    +  2      x    3    −  4      x    2    +  5  x  −  2  ≥  0or      x    5    −  2      x    4    +      x    3    +      x    3    −  2      x    2    +  x  −  2      x    2    +  4  x  −  2  ≥  0or  (  x  −  1      )    2    (      x    3    +  x  −  2  )  ≥  0  ,which is obvious for   x  ≥  1.Done againe!

Gisselle Hodges · Accepted Answer

write your inequality in the form      1    2    +      1          2      (      u      +      1              )        2              ≥      1          1      +      2      u      and square it after this we get  1      /    4                                            u                                3                                    (          5                    u          +          2                                              u                                      2                                +          4          )                                                  (            u            +            1            )                                4                                    (          1          +          2                    u          )                      ≥  0

Prove 1/2 + (1)/(2(u+1)^2) - 1/(sqrt(1+2u)) >= 0 for u>=0

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