Giải thích các bước giải:
$\sqrt{\dfrac{a}{a+2b}}=\dfrac{a\sqrt{3}}{\sqrt{3a(a+2b)}}\ge \dfrac{a\sqrt{3}}{\dfrac{1}{2}(3a+a+2b)}=\dfrac{2\sqrt{3}.a}{4a+2b}$
Chứng minh tương tự ta có :
$\sqrt{\dfrac{b}{b+2c}}\ge \dfrac{2\sqrt{3}.b}{4b+2c}$
$\sqrt{\dfrac{c}{c+2a}}\ge \dfrac{2\sqrt{3}.c}{4c+2a}$
$\rightarrow\sqrt{\dfrac{a}{a+2b}} +\sqrt{\dfrac{b}{b+2c}}+\sqrt{\dfrac{c}{c+2a}}\ge \dfrac{2\sqrt{3}.a}{4a+2b}+\dfrac{2\sqrt{3}.b}{4b+2c}+ \dfrac{2\sqrt{3}.c}{4c+2a}$
$\rightarrow\sqrt{\dfrac{a}{a+2b}} +\sqrt{\dfrac{b}{b+2c}}+\sqrt{\dfrac{c}{c+2a}}\ge 2\sqrt{3}.(\dfrac{a}{4a+2b}+\dfrac{b}{4b+2c}+ \dfrac{c}{4c+2a})$
$\rightarrow\sqrt{\dfrac{a}{a+2b}} +\sqrt{\dfrac{b}{b+2c}}+\sqrt{\dfrac{c}{c+2a}}\ge \sqrt{3}.(\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+ \dfrac{c}{2c+a})$
$\rightarrow\sqrt{\dfrac{a}{a+2b}} +\sqrt{\dfrac{b}{b+2c}}+\sqrt{\dfrac{c}{c+2a}}>\sqrt{3}.(\dfrac{a}{2a+2b+2c}+\dfrac{b}{2b+2c+2a}+ \dfrac{c}{2c+2a+2b})=\dfrac{\sqrt{3}}{2}>1$
