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If $a, b$ are non-negative real numbers, prove that $$ \frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b} $$

I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know $$ \frac{a+b}{1+a+b} \leq \frac{a+b+ab}{1+a+b+ab} $$ which gives me $$ \frac{a}{1+a} + \frac{b}{(1+a)(1+b)} $$

How can I reduce the second term further, and get the required result?

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  • $\begingroup$ I think your second inequality should have $a+b+2ab$ on the numerator of the right hand side? $\frac{a(1+b) + b(1+a)}{(1+a)(1+b)}$. $\endgroup$ – Patrick Stevens Jan 7 '18 at 19:20
  • $\begingroup$ It's possible you intended to assert that $\frac{x}{y} + \frac{m}{n} = \frac{x+m+r}{y+n+r}$, which is very, very untrue in general. $\endgroup$ – Patrick Stevens Jan 7 '18 at 19:21
  • $\begingroup$ Similar question: math.stackexchange.com/questions/2595808/… $\endgroup$ – farruhota Jan 7 '18 at 19:28
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$$ \frac{a+b}{1+a+b} =\frac{a}{1+a+b} + \frac{b}{1+a+b} $$ then prove

$$ \frac{a}{1+a+b} \leq \frac{a}{1+a} $$

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Let $f(x)=\frac{x}{1+x}.$ Then by Jensen inequality $$ f(a)+f(b)=\frac{a}{1+a}+\frac{b}{1+b} \geq 2 f( \frac{a+b}{2})=\frac{a+b}{1+\frac{a+b}{2}} \geq \frac{a+b}{1+a+b}. $$

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We get $$\frac{a}{1+a}+\frac{b}{1+b}-\frac{a+b}{1+a+b}={\frac {ba \left( a+b+2 \right) }{ \left( 1+a \right) \left( 1+b \right) \left( 1+a+b \right) }} \geq 0$$ the numerator can be calculated as $$a(1+b)(1+a+b)+b(1+a)(1+a+b)-(a+b)(1+a)(1+b)=...$$ we also have under the same conditions $$\frac{a+b+c}{1+a+b+c}\le \frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}$$ and so on ...

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In order to prove that $f(x)=\frac{x}{1+x}$ is sublinear on $\mathbb{R}^+$ it is enough to notice that $f'(x)=\frac{1}{(1+x)^2}$ leads to

$$ \frac{f(a+b)-f(a)}{f(b)-f(0)} = \frac{\int_{a}^{a+b}\frac{dx}{(1+x)^2}}{\int_{0}^{b}\frac{dx}{(1+x)^2}}\leq 1 $$ since $f'(x)$ is decreasing. In other terms, the sublinearity is a consequence of the concavity.

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I guess you're almost there...

If $a\ge 0$, then $1+a\ge 1$ and so $(1+a)(1+b)\ge 1+b$. This gives you $$\frac b {(1+a)(1+b)}\le \frac b {1+b}$$ and the result you're looking for follows.

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Since both $a$ and $b$ are nonnegative so are ${{1}\over{1+a}}$ and ${{b}\over{1+b}}+a$ and their multiplication.

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