Inequality with logarithm Hi can you help me with the following
Show that for every $u \ge 0$

$$(1 + u)\log(1 + u) - u \ge \cfrac {u^2} {2 + 2u/3}$$

Obviously i tried to see something clear from Taylor series of $\log(1+x)$ but didn't see something clear.
Thank you!
 A: Reading the book Concentration Inequalities, I found the exact inequality in Ex. 2.8.  I solved it using series expansion (of the $\log$ function) and a carefully chosen geometric series thanks to an inequality proved by induction.  Both are tricky and less straightforward than the direct approach.  I'm posting this for fun and for OP's desire to "see something clear from Taylor series [of the $\log$ function around $1$]".
For any $x>0$, note that $0<x/(1+x)<1$.  The first step allows any "large" $x$ to "fit into the Taylor expansion" by considering its multiplicative reciprocal and inserting a minus sign in front of the $\log$ sign.
$$
\begin{aligned}
& \log(1+x) \\
=& -\log\left(1-\frac{x}{1+x}\right) \\
=& \frac{x}{1+x} + \frac12\left(\frac{x}{1+x}\right)^2 + \frac13\left(\frac{x}{1+x}\right)^3 + \cdots
\end{aligned}
$$
Next, retrieve the LHS by


*

*multiplying $1+x$ on both sides

*moving $x$ from RHS to LHS (with a change of sign)

*on RHS, drag the factor $x$ out of the parenthesis "()" so that it's more pleasing to our eyes.


$$
\begin{aligned}
(1+x)\log(1+x)-x =& x \left[ \frac12\left(\frac{x}{1+x}\right) + \frac13\left(\frac{x}{1+x}\right)^2 + \frac14\left(\frac{x}{1+x}\right)^3 + \cdots \right] \\
=& \frac{x^2}{2(1+x)} \left[ 1 + \frac23\left(\frac{x}{1+x}\right) + \frac24\left(\frac{x}{1+x}\right)^2 + \cdots \right] \\
=& \frac{x^2}{2(1+x)} \sum_{k=0}^\infty \frac{2}{k+2} \left(\frac{x}{1+x}\right)^k
\end{aligned}
$$
The following shows that LHS can be bounded below by a geometric series with common ratio $$\frac23\left(\frac{x}{1+x}\right),$$ so that RHS is shrunk to something simpler than infinite series and more similar to the question.

Lemma     For any non-negative integer $k$, we have
  $$\frac{2}{k+2} \ge \left( \frac23 \right)^k.$$
  Justification by induction
  
  
*
  
*Base case $k=0$: both sides equal to one
  
*Induction hypothesis $k=n$: $$\frac{2}{n+2} \ge \left( \frac23 \right)^n.$$
  
*Inductive step $k=n+1$: $$\frac{2}{n+3} = \frac{2}{n+2} \frac{n+2}{n+3} \ge \left( \frac23 \right)^n \left( 1-\frac{1}{n+3} \right) \ge \left( \frac23 \right)^n \left( 1-\frac13 \right)=\left( \frac23 \right)^{n+1}.\square$$
  

$$
\require{cancel}
\begin{aligned}
(1+x)\log(1+x)-x
=& \frac{x^2}{2(1+x)} \sum_{k=0}^\infty \frac{2}{k+2} \left(\frac{x}{1+x}\right)^k \\
\ge& \frac{x^2}{2(1+x)} \sum_{k=0}^\infty \left( \frac23 \right)^k \left(\frac{x}{1+x}\right)^k \\
=& \frac{x^2}{2(1+x)} \frac{1}{1-\frac23\left(\frac{x}{1+x}\right)} \\
=& \frac{x^2}{2\cancel{(1+x)}} \frac{3\cancel{(1+x)}}{3(1+x)-2x} \\
=& \frac{x^2}{2} \frac{3}{x+3} \\
=& \frac{x^2}{2 \left( 1+x/3 \right)}
\end{aligned}
$$
