<s>[INST] Prove this inequality $\frac{1}{1+a}+\frac{2}{1+a+b}<\sqrt{\frac{1}{a}+\frac{1}{b}}$ 
Let $a,b>0$ show that
  $$\dfrac{1}{1+a}+\dfrac{2}{1+a+b}<\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$$

It suffices to show that
$$\dfrac{(3a+b+3)^2}{((1+a)(1+a+b))^2}<\dfrac{a+b}{ab}$$
or
$$(a+b)[(1+a)(1+a+b)]^2>ab(3a+b+3)^2$$
this idea can't solve it to me,are we aware of an elementary way of proving that? Thanks in advance.
 [/INST] Use Cauchy-Schwarz inequality we have
$$\left(\dfrac{1}{1+a}+\dfrac{2}{1+a+b}\right)^2\le\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}\right)$$
so suffices to show that
$$\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}<1$$
since
$$\dfrac{4b}{(1+a+b)^2}<\dfrac{4b}{4(1+a)b}=\dfrac{1}{1+a}$$
it suffices to show that
$$\dfrac{a}{(1+a)^2}+\dfrac{1}{1+a}<1$$
$$\Longleftrightarrow a+1+a<(1+a)^2\Longleftrightarrow a^2>0$$ it is clear
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