Prove that for all x &gt; 0, 1 + 2 ln ⁡ x ≤ x 2 Prove that for all x &gt; 0 1 + 2 ln ⁡ x ≤ x 2 How can one prove that?

Question

Prove that for all   x  &amp;gt;  0,   1  +  2  ln  ⁡  x  ≤      x    2  Prove that for all   x  &amp;gt;  0  1  +  2  ln  ⁡  x  ≤      x    2  How can one prove that?

Marlee Norman · Accepted Answer

First, note the problem cannot be for all     x  ∈      R     but, at most , for     x  &amp;gt;  0    (why?). Now, a simple calculus approach: define  f  (  x  )  :=  1  +  2  log  ⁡  x  −      x    2      ⟹        f    ′    (  x  )  =      2    x    −  2  x  =  0    ⟺    x  =  1and since         f    ″    (  1  )  =  −  2  &amp;lt;  0   , we get a maximum point at     (  1  ,  0  )   , meaning that for all     x  &amp;lt;  0   we have  f  (  x  )  ≤  0

Hailie Blevins · Accepted Answer

Notice by the famous inequality       e    x    ≥  1  +  x. We can rescale this and obtain      e          x      −      1        ≥  x    ⟹    x  −  1  ≥  ln  ⁡  x    ⟹    2  x  −  2  ≥  2  ln  ⁡  x    ⟹    2  x  −  1  ≥  2  ln  ⁡  x  +  1Now, by using             a      +      b        2    ≥      a    b   with   a  =      x    2   and   b  =  1, we obtain                    x        2            +      1        2    ≥  x    ⟹        x    2    ≥  2  x  −  1Now, by transitivity,      x    2    ≥  2  ln  ⁡  x  +  1QEDActually a shorter solution:Since       e    x    ≥  1  +  x, then       e                  x        2              ≥  1  +      x    2  . Translate this to obtain      e                  x        2            −      1        ≥      x    2      ⟹        x    2    −  1  ≥  ln  ⁡      x    2      ⟹        x    2    ≥  1  +  2  ln  ⁡  x

Prove that for all x > 0 , 1 + 2 ln ⁡ x ≤

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