Toán Lớp 9: Chứng minh với a,b,c>0
∑ $\frac{a²-bc}{2a²+b²+c²}$ ≥0
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Toán Lớp 9: Chứng minh với a,b,c>0
∑ $\frac{a²-bc}{2a²+b²+c²}$ ≥0
Comments ( 1 )
Lời giải và giải thích chi tiết:
$->\text{Áp dụng bất đẳng thức}$ $Co-si$:
$$=> – \sum \dfrac{bc}{2a^2+b^2+c^2} \ge -(\dfrac{b^2+c^2}{4a^2+2b^2+2c^2}+\dfrac{a^2+b^2}{4c^2+2a^2+2b^2}+\dfrac{a^2+c^2}{4b^2+a^2+c^2}) = \sum -[\dfrac{a^2}{2}.(\dfrac{1}{2b^2+a^2+c^2}+\dfrac{1}{2c^2+a^2+b^2}]$$
$->\text{Áp dụng bổ đề:}$ $\dfrac{1}{x}+\dfrac{1}{y} \ge \dfrac{4}{x+y}$
$$=>VT \ge \sum -[\dfrac{a^2}{8}.(\dfrac{1}{b^2+a^2}+\dfrac{1}{b^2+c^2} +\dfrac{1}{c^2+a^2}+\dfrac{1}{c^2+b^2}] \ge \sum -[\dfrac{a^2}{32}.(\dfrac{1}{b^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{1}{c^2}+\dfrac{1}{a^2}+\dfrac{1}{c^2}+\dfrac{1}{b^2}) = \sum -[\dfrac{a^2}{32}.(\dfrac{2}{a^2}+\dfrac{3}{b^2}+\dfrac{3}{c^2})]$$
$\text{Quay lại, ta áp dụng vào bài toán:}$
$$=>\sum \dfrac{a^2-bc}{2a^2+b^2+c^2} \ge \sum [\dfrac{a^2}{2a^2+b^2+c^2} -\dfrac{a^2}{32}.(\dfrac{2}{a^2}+\dfrac{3}{b^2}+\dfrac{3}{c^2})] = \sum [\dfrac{1}{2+\dfrac{b^2}{a^2}+\dfrac{c^2}{a^2}} – \dfrac{1}{32}.(2+\dfrac{3a^2}{b^2}+\dfrac{3a^2}{c^2})] (”*”)$$
$\text{Đặt}$: \(\left[ \begin{array}{l}x=\dfrac{a^2}{b^2}\\y=\dfrac{b^2}{c^2} \\ z = \dfrac{c^2}{a^2} \\ x,y,z >0\end{array} \right.\)
$=>$ $\text{Biểu thức}$ $(”*”)$ $\text{trở thành:}$ $\sum [\dfrac{1}{2+\dfrac{1}{x}+z} – \dfrac{1}{32}.(2+3x+\dfrac{3}{z})] $
$\text{Áp dụng bất đẳng thức}$ $Svac-xơ$:
$$=>VT \ge \dfrac{9}{x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+6} – \dfrac{3}{32}.(2+x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})$$
$\text{Đặt}$: $x+y+z+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}= t$ $(t \ge 6)$
$$=> VT \ge \dfrac{9}{t+6}-\dfrac{3}{32}.(2+t)= \dfrac{9}{t+6}-\dfrac{3}{32}.t-\dfrac{3}{16}=3.(\dfrac{3}{t+6}-\dfrac{1}{32}.t-\dfrac{2}{32})= 3.(\dfrac{-t^2-8t+84}{32.(t+6)} )= 3.\dfrac{(t-6)(t+14)}{32.(t+6)} \ge 0$$
$$\text{(Luôn đúng, Điều phải chứng minh)}$$
$\text{Dấu bằng xảy ra khi:}$ $t=6 =>x=y=z => a=b=c$
*Do bạn thích dùng $\sum_{cyc}$ nên hình cũng dùng theo cho bài đỡ dài :>