$\{\begin{array}{c}\frac{A}{6}=\frac{B}{2}=\frac{c}{1}=\frac{A+B+C}{6+2+1}=\frac{180}{9}=20\\ A=120\\ B=40\\ C=20\end{array}$

$\{\begin{array}{c}\widehat{BAD}={60}^0>\hat{A}\end{array}$

B

B

Bunhiacopxki:

\(\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)\ge\left(ab+bc+ca\right)^2\)

\(\Rightarrow\dfrac{ab}{a^2+bc+ca}\le\dfrac{ab\left(b^2+bc+ca\right)}{\left(ab+bc+ca\right)^2}\)

Tương tự: \(\dfrac{bc}{b^2+ca+ab}\le\dfrac{bc\left(c^2+ca+ab\right)}{\left(ab+bc+ca\right)^2}\)

\(\dfrac{ca}{c^2+ab+bc}\le\dfrac{ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

\(\Rightarrow VT\le\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\)

Nên ta chỉ cần chứng minh:

\(\dfrac{ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)}{\left(ab+bc+ca\right)^2}\le\dfrac{a^2+c^2+c^2}{ab+bc+ca}\)

\(\Leftrightarrow ab\left(b^2+bc+ca\right)+bc\left(c^2+ca+ab\right)+ca\left(a^2+ab+bc\right)\le\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)\)

Nhân phá và rút gọn 2 vế:

\(\Leftrightarrow a^3b+b^3c+c^3a\ge abc\left(a+b+c\right)\)

\(\Leftrightarrow\dfrac{a^3b+b^3c+c^3a}{abc}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)

Đúng do: \(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c}=a+b+c\)

Dấu "=" xảy ra khi \(a=b=c\)

\(VT=\dfrac{1}{\dfrac{a}{b}+\dfrac{c}{a}+1}+\dfrac{1}{\dfrac{b}{c}+\dfrac{a}{b}+1}+\dfrac{1}{\dfrac{c}{a}+\dfrac{b}{c}+1}\)

\(\left(\dfrac{a}{b},\dfrac{b}{c},\dfrac{c}{a}\right)\rightarrow\left(x^3,y^3,z^3\right)\)\(\Rightarrow xyz=1\).

\(VT=\sum\dfrac{1}{x^3+y^3+1}\le\sum\dfrac{1}{xy\left(x+y\right)+xyz}=\sum\dfrac{z}{x+y+z}=1\)

Dấu = xảy ra khi x=y=z=1 hay a=b=c

Hay quá bạn ơi tks

\(BĐT\Leftrightarrow\)∑\(\left(\frac{b^2}{c}+a+b\right)\)\(\le\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a+b+c\le\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b}\)

\(\Leftrightarrow\frac{\left(a-c\right)^2}{c}+\frac{\left(b-a\right)^2}{a}+\frac{\left(c-b\right)^2}{b}\ge0\)

Bạn tham khảo lời giải tại đây:

Câu hỏi của Nguyễn Xuân Đình Lực - Toán lớp 9 | Học trực tuyến