Áp dụng BĐT Schwarz ta có:
\(\sqrt{a^2+3b^2}=\sqrt{a^2+b^2+b^2+b^2}\ge\sqrt{\frac{\left(a+b+b+b\right)^2}{4}}=\sqrt{\frac{\left(a+3b\right)^2}{4}}\)
Chứng minh tương tự ta có:
\(\sqrt{3a^2+b^2}\ge\sqrt{\frac{\left(3a+b\right)^2}{4}}\)
Như vậy ta có:
\(\frac{a+2b}{\sqrt{3a^2+b^2}+\sqrt{a^2+3b^2}+2b}\le\frac{a+2b}{\sqrt{\frac{\left(3a+b\right)^2}{4}}+\sqrt{\frac{\left(a+3b\right)^2}{4}}+2b}=\frac{a+2b}{\frac{3a+b}{2}+\frac{3b+a}{2}+2b}=\frac{a+2b}{2a+2b+2b}=\frac{a+2b}{2a+4b}=\frac{a+2b}{2\left(a+2b\right)}=\frac{1}{2}\)
Dấu \("="\) xảy ra \(\Leftrightarrow a=b\)
(Không chắc lắm ạ!)
