Đặt \(\left\{{}\begin{matrix}b+c-a=x\\c+a-b=y\\a+b-c=z\end{matrix}\right.\)\(\left(x,y,z>0\right)\)\(\Rightarrow\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\x+z=2b\end{matrix}\right.\)

Thì ta có: \(\dfrac{2\left(y+z\right)}{x}+\dfrac{9\left(x+z\right)}{2y}+\dfrac{8\left(x+y\right)}{z}\ge26\)

Áp dụng BĐT AM-GM ta có:

\(VT=\dfrac{2\left(y+z\right)}{x}+\dfrac{9\left(x+z\right)}{2y}+\dfrac{8\left(x+y\right)}{z}\)

\(=\dfrac{2y}{x}+\dfrac{2z}{x}+\dfrac{9x}{2y}+\dfrac{9z}{2y}+\dfrac{8x}{z}+\dfrac{8y}{z}\)

\(=\left(\dfrac{2y}{x}+\dfrac{9x}{2y}\right)+\left(\dfrac{2z}{x}+\dfrac{8x}{z}\right)+\left(\dfrac{9z}{2y}+\dfrac{8y}{z}\right)\)

\(\ge2\sqrt{\dfrac{2y}{x}\cdot\dfrac{9x}{2y}}+2\sqrt{\dfrac{2z}{x}\cdot\dfrac{8x}{z}}+2\sqrt{\dfrac{9z}{2y}\cdot\dfrac{8y}{z}}\)

\(\ge6+8+12=26=VP\)

Lời giải:

Áp dụng BĐT Cauchy-Schwarz ta có:
\(P=\frac{4a}{b+c-a}+\frac{9b}{a+c-b}+\frac{16c}{a+b-c}\)

\(P+\frac{29}{2}=\frac{4a}{b+c-a}+2+\frac{9b}{a+c-b}+\frac{9}{2}+\frac{16c}{a+b-c}+8\)

\(=\frac{2(a+b+c)}{b+c-a}+\frac{9(a+b+c)}{2(a+c-b)}+\frac{8(a+b+c)}{a+b-c}\)

\(=2(a+b+c)\left(\frac{1}{b+c-a}+\frac{\frac{9}{4}}{a+c-b}+\frac{4}{a+b-c}\right)\)

\(\geq 2(a+b+c).\frac{(1+\frac{3}{2}+2)^2}{b+c-a+a+c-b+a+b-c}=\frac{81}{2}.(a+b+c).\frac{1}{a+b+c}=\frac{81}{2}\)

\(\Rightarrow P\geq \frac{81}{2}-\frac{29}{2}=26\)

Vậy \(P_{\min}=26\)

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Câu hỏi của Trương quang huy hoàng - Toán lớp 9 | Học trực tuyến

Lời giải:

Ta có:

\(A=\frac{4a}{b+c-a}+\frac{9b}{a+c-b}+\frac{16c}{a+b-c}\)

\(\Rightarrow A+\frac{29}{2}=\frac{4a}{b+c-a}+2+\frac{9b}{a+c-b}+\frac{9}{2}+\frac{16c}{a+b-c}+8\)

\(A+\frac{29}{2}=\frac{2(a+b+c)}{b+c-a}+\frac{\frac{9}{2}(a+b+c)}{a+c-b}+\frac{8(a+b+c)}{a+b-c}\)

\(A+\frac{29}{2}=(a+b+c)\left(\frac{2}{b+c-a}+\frac{\frac{9}{2}}{a+c-b}+\frac{8}{a+b-c}\right)\)

\(\geq (a+b+c).\frac{(\sqrt{2}+\sqrt{\frac{9}{2}}+\sqrt{8})^2}{b+c-a+a+c-b+a+b-c}=\frac{81}{2}\)

(Áp dụng BĐT S.Vac -xơ)

\(\Rightarrow A\geq 26\)

Vậy \(A_{\min}=26\)

: -> Câu hỏi của Almira

Đặt \(b+c-a=2x,c+a-b=2y,a+b-c=2z\to x,y,z>0\)  v

à thỏa mãn \(a=y+z,b=z+x,c=x+y.\) Đặt \(S=2VT\)  (hai lần vế trái của bất đẳng thức)  thì ta có

\(S=\frac{4\left(y+z\right)}{x}+\frac{9\left(x+z\right)}{y}+\frac{16\left(x+y\right)}{z}=\left(\frac{4y}{x}+\frac{9x}{y}\right)+\left(\frac{4z}{x}+\frac{16x}{z}\right)+\left(\frac{9z}{y}+\frac{16y}{z}\right)\)

Theo bất đẳng thức Cô-Si ta được

\(S\ge2\sqrt{\frac{4y}{x}\cdot\frac{9x}{y}}+2\sqrt{\frac{4z}{x}\cdot\frac{16x}{z}}+2\sqrt{\frac{9z}{y}\cdot\frac{16y}{z}}=2\cdot6+2\cdot8+2\cdot12=2\cdot26=52.\)

Suy ra \(VT=\frac{S}{2}\ge\frac{52}{2}=26\).   (ĐPCM)

đề sai ở mẫu cuối nhé

đặt b + c - a = x ; a + c - b = y ; a + b - c = z

\(\Rightarrow a=\frac{y+z}{2};b=\frac{x+z}{2};c=\frac{x+y}{2}\)

\(\Rightarrow P=\frac{2\left(y+z\right)}{x}+\frac{9\left(x+z\right)}{2y}+\frac{8\left(x+y\right)}{z}=\frac{2y}{x}+\frac{9x}{2y}+\frac{2z}{x}+\frac{8x}{z}+\frac{9z}{2y}+\frac{8y}{z}\)

\(\ge6+8+12=26\)

bài này dấu ' =" giải ra mệt lắm nên bạn tự giải

\(A=\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}\ge\dfrac{4}{a+b-c+b+c-a}\ge\dfrac{4}{2b}\ge\dfrac{2}{b}\\\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\ge\dfrac{4}{b+c-a+c+a-b}\ge\dfrac{4}{2c}\ge\dfrac{2}{c}\\\dfrac{1}{a+b-c}+\dfrac{1}{c+a-b}\ge\dfrac{4}{a+b-c+c+a-b}\ge\dfrac{4}{2a}\ge\dfrac{2}{a}\end{matrix}\right.\)

\(\Rightarrow2\left(\dfrac{1}{a+b-c}+\dfrac{1}{b+c-a}+\dfrac{1}{c+a-b}\right)\ge\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Rightarrow A\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) \(dấu"="xảy\) \(ra\Leftrightarrow a=b=c\)

đề có sai 1 chút nha bạn :

đề phải là \(a;b;c>0\) : \(CMR\) \(\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\ge6\) mới đúng

giải

đặt : \(P=\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\)

ta có : \(P=\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\)

\(P=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{9b}{a+c}+9\right)+\left(\dfrac{16c}{a+b}+16\right)-26\)

\(P=\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)-26\)

\(P=\dfrac{1}{2}\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)-26\)

áp dụng bất đẳng thức Bunhiacopxki

ta có :

\(\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)\ge\left(\sqrt{1}+\sqrt{9}+\sqrt{16}\right)^2\)

\(\Leftrightarrow\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)\ge64\)

\(\Leftrightarrow\) \(P=\dfrac{1}{2}\left(\left(b+c\right)+\left(a+c\right)+\left(a+b\right)\right)\left(\dfrac{1}{b+c}+\dfrac{9}{a+c}+\dfrac{16}{a+b}\right)-26\ge\dfrac{1}{2}.64-26\)

vậy \(P=\dfrac{a}{b+c}+\dfrac{9b}{a+c}+\dfrac{16c}{a+b}\ge6\) (đpcm)

dấu "=" xảy ra khi \(b+c=\dfrac{a+c}{9}=\dfrac{a+b}{16}\)