How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$?

$$ \frac{a+b}{1+a+b} =\frac{a}{1+a+b} + \frac{b}{1+a+b} $$ then prove

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

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 ...

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