Cho a, b, c là các số thực không âm, không có hai số nào đồng thời bằng không. Chứng minh rằng:
$\sum \dfrac{a^2}{(b^2+bc+c^2)^2}+\dfrac{6(ab+ac+bc)}{(a+b+c)^4}\geq \dfrac{5}{(a+b+c)^2}$
Mình mạn phép chém bài này nha
trước hết ta chứng minh
$\dfrac{a}{{{b^2} + bc + {c^2}}} + \dfrac{b}{{{c^2} + ac + {a^2}}} + \dfrac{c}{{{a^2} + ab + {b^2}}} \ge \dfrac{{(a + b + c)}}{{(ab + bc + ca)}}$
thật vậy
$\dfrac{a}{{{b^2} + bc + {c^2}}} = \dfrac{{4a(ab + bc + ca)}}{{4(ab + bc + ca)({b^2} + bc + {c^2})}} \ge \dfrac{{4a(ab + bc + ca)}}{{{{(b + c)}^2}{{(a + b + c)}^2}}} \Rightarrow \dfrac{a}{{{{(b + c)}^2}{{(a + b + c)}^2}}} + \dfrac{b}{{{{(c + a)}^2}{{(a + b + c)}^2}}} + \dfrac{c}{{{{(a + b)}^2}{{(a + b + c)}^2}}} \ge \dfrac{{(a + b + c)}}{{4{{(ab + bc + ca)}^2}}} \Leftrightarrow \dfrac{a}{{{{(b + c)}^2}}} + \dfrac{b}{{{{(c + a)}^2}}} + \dfrac{c}{{{{(a + b)}^2}}} \ge \dfrac{{{{(a + b + c)}^3}}}{{4{{(ab + bc + ca)}^2}}}$
Mà theo BCS thì
$ \Leftrightarrow (\dfrac{a}{{{{(b + c)}^2}}} + \dfrac{b}{{{{(c + a)}^2}}} + \dfrac{c}{{{{(a + b)}^2}}})(a + b + c) \ge {\left( {\dfrac{a}{{b + c}} + \dfrac{b}{{a + c}} + \dfrac{c}{{a + b}}} \right)^2} \ge \dfrac{{{{(a + b + c)}^4}}}{{4{{(ab + bc + ca)}^2}}}$
vậy
$\dfrac{a}{{{{(b + c)}^2}}} + \dfrac{b}{{{{(c + a)}^2}}} + \dfrac{c}{{{{(a + b)}^2}}} \ge \dfrac{{{{(a + b + c)}^3}}}{{4{{(ab + bc + ca)}^2}}}$
bổ đề dc cm
dễ thấy
$\dfrac{{{a^2}}}{{{{({b^2} + bc + {c^2})}^2}}} + \dfrac{{{b^2}}}{{{{({a^2} + ac + {c^2})}^2}}} + \dfrac{{{c^2}}}{{{{({a^2} + ab + {b^2})}^2}}} \ge \dfrac{1}{3}{(\dfrac{{a + b + c}}{{ab + bc + ca}})^2}$
Giả sử ab+bc+ca=1áp dụng AM GM ta có
$\dfrac{1}{3}{(a + b + c)^2} + \dfrac{6}{{{{(a + b + c)}^4}}} \ge 5\sqrt[5]{{\dfrac{{{{(a + b + c)}^6}}}{{{9^3}}}\dfrac{{{3^2}}}{{{{(a + b + c)}^8}}}}} \ge VP$
vậy ta có ĐPCM