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How to show $\frac{2x}{2+x}<\log(x+1)$ for $x>0$

Without differentiating, more elementary (it looks then more complicated but OK)

$\log(x+1)=\int\limits_1^{x+1}\frac 1udu$ and $\frac{2x}{2+x}=x\frac{1}{ 1+\frac x2}$ hence

$\log(x+1)-\frac{2x}{2+x}=\int\limits_1^{x+1}\frac 1u-\frac{1}{ 1+\frac x2}du=\int\limits_1^{x+1}\frac{1+\frac x2-u}{u\left(1+\frac x2\right)}du=\int\limits_1^{\frac x2+1}\frac{1+\frac x2-u}{u\left(1+\frac x2\right)}du-\int\limits_{\frac x2+1}^{x+1}\frac{u-\left(1+\frac x2\right)}{u\left(1+\frac x2\right)}du$

The numerator of the integrand in the first integral is positive and negative in the $2$nd. But for the first one the denominatior is smaller, so the fraction is always bigger than its corresponding fraction in the second one, and so the difference is always positive is this correct, do you have an alternative proof

user257
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2 Answers2

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Following your first idea, you have $$\frac{2x}{2+x}=\int_1^{1+x}\frac{4}{(u+1)^2} du$$ I think that you can take it from here ....

Kelenner
  • 18,894
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Note that $$ \frac{4}{(x+2)^2}-\frac{1}{x+1}=-\frac{x^2}{(x+1)(x+2)^2}<0$$ and hence $$ \frac{4}{(x+2)^2}<\frac{1}{x+1}. $$ Integrating from $0$ to $x$ ($x>0$), one has $$ \int_0^x\frac{4}{(t+2)^2}dt<\int_0^x\frac{1}{t+1}dt $$ which gives $$ \frac{2x}{2+x}<\ln(x+1). $$ Done.

xpaul
  • 47,821