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The following inequality is valid for all positive real $x$, $$ x - (1+x)\log(1+x) \leq \frac{-x^2}{2(1+x)}. $$ It is possible to show that this is true by considering the function $$ f(x) := x - (1+x)\log(1+x)+ \frac{x^2}{2(1+x)}. $$ By differentiation it is possible to check that $f(x)$ attains it maximum on the nonnegative reals at $x = 0$.

However, is there a cleaner, more obvious way to see that this is true? Specifically, I would like a solution that does not require me to analyze the monotonicity of $f$ via differentiation if possible. I tried Taylor expansion, but do not get this inequality.

Angelo
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Drew Brady
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    I have not analyzed yet, but the second answer there doesn't seem to use differentiation (but integration) – VIVID Oct 02 '20 at 17:15
  • One could accept $$\log(x) \leq \frac{(x-1)(x+5)}{4x+2}$$ as obvious since it's quite famous. But OK, then let's wait for any better answer. – VIVID Oct 02 '20 at 17:23
  • I don't know if that has a name, but if you search for that, you find it's often used as a known fact. – VIVID Oct 02 '20 at 17:32
  • I agree with Drew Brady completely. – Angelo Oct 02 '20 at 18:36
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    $$\log(1+y)\le y-\dfrac{1}{2}y^2\quad\text{ for all };y\in\left]-1,0\right];.\quad\color{blue}{(*)}$$

    For all $;x\ge0;,;$ it results that $;y=\dfrac{1}{1+x}-1\in\left]-1,0\right],$ hence, by applying $(*)$, we get that

    $$\log\left(1+\dfrac{1}{1+x}-1\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\left(\dfrac{1}{1+x}-1\right)^2,$$

    $$-\log\left(1+x\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\dfrac{x^2}{(1+x)^2};,$$

    $$-(1+x)\log\left(1+x\right)\le 1-(1+x)-\dfrac{x^2}{2(1+x)};,$$

    $$-(1+x)\log\left(1+x\right)\le -x-\dfrac{x^2}{2(1+x)};,$$

    $$x-(1+x)\log(1+x)+\dfrac{x^2}{2(1+x)}\le0;.$$

     – Angelo Oct 02 '20 at 18:39
  • Sorry, I voted to close while adding my very first comment. I have voted to reopen now. – VIVID Oct 02 '20 at 18:56
  • Also, Angelo could post the answer there in order to protect MSE from very similar questions. – VIVID Oct 02 '20 at 19:02
  • @DrewBrady, you are welcome. – Angelo Oct 02 '20 at 19:10
  • By the way, in your approach via differentiation, it is simpler to consider $f(x) := \frac{x}{1+x} - \log (1+x) + \frac{x^2}{2(1+x)^2}$. Then $f'(x) = - \frac{x^2}{(1+x)^3} < 0$. – River Li Oct 03 '20 at 03:55

2 Answers2

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Better proof:

$\log(1+y)\le y-\dfrac{1}{2}y^2\quad$ for all $\;y\in\left]-1,0\right]\;.\quad\color{blue}{(*)}$

For all $\;x\ge0\;,\;$ it results that $\;y=\dfrac{1}{1+x}-1\in\left]-1,0\right],$ hence, by applying $(*)$, we get that

$\log\left(1+\dfrac{1}{1+x}-1\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\left(\dfrac{1}{1+x}-1\right)^2,$

$-\log\left(1+x\right)\le\dfrac{1}{1+x}-1-\dfrac{1}{2}\dfrac{x^2}{(1+x)^2}\;,$

$-(1+x)\log\left(1+x\right)\le 1-(1+x)-\dfrac{x^2}{2(1+x)}\;,$

$-(1+x)\log\left(1+x\right)\le -x-\dfrac{x^2}{2(1+x)}\;,$

$x-(1+x)\log(1+x)+\dfrac{x^2}{2(1+x)}\le0\;,\;$ for all $\;x\ge0\;.$

Angelo
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Yet another way to obtain such a bound is as follows. Set $\phi(\delta) = (1+\delta)\log(1+\delta)$. One then checks that $$ \phi(0) = 0, \quad \dot{\phi}(0) = 1, \quad \mbox{and} \quad \ddot{\phi}(z) = \frac{1}{1+z}. $$ Therefore, by Taylor's theorem, for some $\tilde x \in (0, x)$: $$ x - (1+x)\log(1+x) = x - \phi(x) = x - \Big\{\phi(0) + \dot{\phi}(0) x + \ddot{\phi}(\tilde x) \frac{x^2}{2}\Big\} \leq \frac{-x^2}{2(1+x)} $$ as claimed. (The final inequality uses $\ddot{\phi}(\tilde x) \geq \tfrac{1}{1 + x}$.)

Drew Brady
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