Prove the value of $\ln\frac{s + x}{x} - \frac{s}{(s + x)}$ is non-negative

Question

I have the following function and I am looking to prove it is non-negative.

$f(s,x)=\ln\frac{s + x}{x} - \frac{s}{(s + x)}$, for $x$ and $s$ are both positive.

To prove $f(s,x) \geq 0 $

I tried some numerical values. It shows that the minimum value is $0$.

But I don't know how I can prove it is non-negative.

Thanks!

Answer 1

Denote: $\frac sx=t\ge 0$. Then: $$f(s,x)=\ln\frac{s + x}{x} - \frac{s}{(s + x)}=\ln \left(\frac{s}{x}+1\right)-\frac{1}{1+\frac xs}\\ f(t)=\ln(t+1)-\frac{1}{1+\frac 1t}=\ln (t+1)-\frac{t}{1+t}.$$ Note: $$f(0)=0\\ f'(t)=\frac{1}{1+t}-\frac{1}{(1+t)^2}=\frac{t}{(1+t)^2}>0,t>0$$ So, $f(t)$ is increasing at $t>0$ and $f(0)=0$ is the minimum.

Answer 2

Hint: It suffices to show that $\ln x\ge 1-\frac{1}{x}$ for all $x>0$, or equivalently, $g(x)=x\ln x-x\ge -1$. Try to find the minimum of $g$ on $(0,\infty)$.

Answer 3

So if $s >0$ and $x > 0$ as you mentioned in the comment above. $$f(s,x) = \ln \frac{s+x}{x} - \frac{s}{s+x} = \ln \frac{s+x}{x} - \frac{s+x-x}{s+x} = \ln \frac{s+x}{x} - 1 + \frac{x}{s+x}$$ Let's assume $z = \frac{s+x}{x}$, then $$f(z) = \ln z - 1 + \frac{1}{z}$$ is your function. Setting the derivative to zero, we get $$f'(z) = \frac{1}{z} - \frac{1}{z^2} =0$$Hence at $z = 1$ we have a minimum (could be shown from $2^{nd}$ derivative), then $f(1) = 0$ is the minimum.

Answer 4

Rewriting as:

$$ f(s,x) = \left[\ln \left(1+\frac{s}{x} \right)-1 \right]+\frac{1}{1+\frac{s}{x}} $$

Then we can reformulate the problem as a single variable optimization: $$ y=1+\frac{s}{x}> 1 $$ $$ g(y) = \ln(y) -1 +\frac{1}{y} $$

We know that $g(1)=0$, and that: $$ g'(y) = \frac{1}{y}-\frac{1}{y^2}=\frac{1}{y} \left( 1-\frac{1}{y}\right) $$ So $g'(1)=0$ but for any $y>1$ we have $g'(y)>0$, therefore $g$ is monotonically increasing, and therefore non-negative.

Answer 5

Let x,s >0, real.

MVT

$I:=\log {\displaystyle \int_{x}^{s+x}}(1/t)dt=$

$(1/r)((s+x)-(x))= (1/r)s$, where $r \in [x,x+s]$.

Hence :

$f(x,s):= I-\dfrac{s}{s+x} =$

$\dfrac{s}{r} - \dfrac{s}{s+x} \ge$

$\dfrac{s}{s+x}-\dfrac{s}{s+x}=0$.