a/(1+a) +2b/(1+b)=1 ->( a(1+b) + 2b(1+a) )/(1+a)(1+b)=1 -> a(1+b) + 2b(1+a)=(1+a)(1+b) -> a+ab+2b+2ab=1+a+b+ab -> b+2ab=1
Áp dụng BĐT Cauchy ta có:
[tex]b+2ab \geq 2\sqrt{b.2ab}=2\sqrt{2ab^2}[/tex]
[tex]\rightarrow 1\geq 2\sqrt{2ab^2} \rightarrow \sqrt{2ab^2}\leq \frac{1}{2} \rightarrow 2ab^2\leq \frac{1}{4}\rightarrow ab^2\leq \frac{1}{8} (dpcm)[/tex]