Áp dụng BĐT Cô-si cho các số dương ta có:
(2a+b+c)2 = \(\left[\left(a+b\right)+\left(a+c\right)\right]^2\) \(\ge\) 4(a+b)(a+c)
\(\Rightarrow\) \(\dfrac{1}{\left(2a+b+c\right)^2}\) \(\le\) \(\dfrac{1}{4\left(a+b\right)\left(a+c\right)}\)
Tương tự : \(\dfrac{1}{\left(2b+c+a\right)^2}\) \(\le\) \(\dfrac{1}{4\left(b+c\right)\left(b+a\right)}\)
\(\dfrac{1}{\left(2c+a+b\right)^2}\) \(\le\) \(\dfrac{1}{4\left(c+b\right)\left(c+a\right)}\)
Cộng theo vế 3 đẳng thức trên
\(\dfrac{1}{\left(2a+b+c\right)^2}\)+\(\dfrac{1}{\left(2b+c+a\right)^2}\)+\(\dfrac{1}{\left(2c+a+b\right)^2}\) \(\le\)\(\dfrac{1}{4}\left(\dfrac{1}{\left(a+b\right)\left(a+c\right)}+\dfrac{1}{\left(b+c\right)\left(b+a\right)}+\dfrac{1}{\left(c+b\right)\left(c+a\right)}\right)\)
=\(\dfrac{1}{4}\left(\dfrac{b+c+a+b+c+a}{\left(a+b\right)\left(a+c\right)\left(b+c\right)}\right)\)
=\(\dfrac{1}{2}\left(\dfrac{a+b+c}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\right)\)
Áp dụng BĐT Cô-si ta có:
\(a+b\ge2\sqrt{ab}\)
\(b+c\ge2\sqrt{bc}\)
\(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\) \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\)
\(\Rightarrow\) P \(\le\) \(\dfrac{a+b+c}{16abc}\) = \(\dfrac{1}{16}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)\) \(\le16\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\) = \(\dfrac{3}{16}\)
\(\Rightarrow\) Pmax = \(\dfrac{3}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow\) a = b = c = 1
Vậy Pmax = \(\dfrac{3}{16}\) \(\Leftrightarrow\) a = b = c = 1
