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.
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
if op go ahead , he will get from his last step :LHS-RHS$=(4a^4+4a^2+b^3-6a^2b)+(a^2b^2-3ab+b+a)+a^2b^3+ab^3+3a^3b^2+a^2b^2+ab^2+2b^2+3a^4b+a^3b+a^5+6a^3 >0 \iff$
$(4a^4+4a^2+b^3-6a^2b) \ge 0 \cap (a^2b^2-3ab+b+a)\ge 0$
