\(\Sigma\dfrac{a}{\sqrt{2a+b}}=\Sigma\dfrac{a}{\sqrt{\left(a+b+c\right)\left(2a+b\right)}}\le\dfrac{\dfrac{a}{a+b+c}+\dfrac{a}{2a+b}+\dfrac{b}{a+b+c}+\dfrac{b}{2b+c}+\dfrac{c}{a+b+c}+\dfrac{c}{2c+a}}{2}=\dfrac{1+\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}}{2}\)
\(ta\) \(đi\) \(cminh:\) \(\dfrac{a}{2a+b}+\dfrac{b}{2b+c}+\dfrac{c}{2c+a}\le1\left(1\right)\)
\(\left(1\right)\Leftrightarrow\left(2a+b\right)\left(2b+c\right)\left(2c+a\right)\ge a\left(2b+c\right)\left(2c+a\right)+b\left(2a+b\right)\left(2c+a\right)+c\left(2a+b\right)\left(2b+c\right)\)
\(\Leftrightarrow a^2c+b^2a+bc^2\ge3abc\)(đúng theo cô si 3 số)
do đó \(\left(1\right)\)đúng \(\Rightarrow\Sigma\dfrac{a}{\sqrt{2a+b}}\le\dfrac{1+1}{2}=1\)
