$\dfrac{b.\sqrt[]b}{\sqrt[]{2a+b+c}}$
$=\dfrac{b}{\sqrt[]{\dfrac{2a+b+c}{b}}}$
$=\dfrac{b}{\sqrt[]{\dfrac{a+3}{b}}}$
Áp dụng bất đẳng thức Cauchy có:
$\dfrac{1}{2}.\sqrt[]{4.\dfrac{a+3}{b}}≤\dfrac{1}{2}.\dfrac{4+\dfrac{a+3}{b}}{2}=\dfrac{a+4b+3}{4b}$
$⇒\dfrac{b}{\sqrt[]{\dfrac{a+3}{b}}}≥\dfrac{4b^2}{a+4b+3}$
Hay $\dfrac{b.\sqrt[]b}{\sqrt[]{2a+b+c}}≥\dfrac{4b^2}{a+4b+3}$
Tương tự: $\dfrac{c.\sqrt[]c}{\sqrt[]{2b+c+a}}≥\dfrac{4c^2}{b+4c+3}$
$\dfrac{a.\sqrt[]a}{\sqrt[]{2c+a+b}}≥\dfrac{4a^2}{c+4a+3}$
Nên $P≥\dfrac{4b^2}{a+4b+3}+\dfrac{4c^2}{b+4c+3}+\dfrac{4a^2}{c+4a+3}$
Áp dụng bất đẳng thức Svacxo có:
$\dfrac{4b^2}{a+4b+3}+\dfrac{4c^2}{b+4c+3}+\dfrac{4a^2}{c+4a+3}≥4.\dfrac{(a+b+c)^2}{5.(a+b+c)+9}=4.\dfrac{3^2}{5.3+9}=\dfrac{3}{2}$
Hay $P≥\dfrac{3}{2}$
Dấu $=$ xảy ra $⇔a=b=c=1$
