Áp dụng liên tiếp Bunyakovsky và Cauchy-Schwarz ta được: \(NL^2=\left(\sqrt{\dfrac{a}{2a+b+c}}+\sqrt{\dfrac{b}{2b+a+c}}+\sqrt{\dfrac{c}{2c+a+b}}\right)^2\) \(\le\left(1^2+1^2+1^2\right)\left(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\right)=3\left(\dfrac{a}{2a+b+c}+\dfrac{b}{2b+a+c}+\dfrac{c}{2c+a+b}\right)\) \(=3\left(\dfrac{a}{\left(a+b\right)+\left(a+c\right)}+\dfrac{b}{\left(a+b\right)+\left(b+c\right)}+\dfrac{c}{\left(a+c\right)+\left(b+c\right)}\right)\le\dfrac{3}{4}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}+\dfrac{c}{b+c}\right)=\dfrac{3}{4}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\right)=\dfrac{3}{4}.3=\dfrac{9}{4}\)\(NL^2\le\dfrac{9}{4}\Leftrightarrow NL\le\dfrac{3}{2}\).Dấu "=" khi \(a=b=c\)
