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

đặt \(b+c-a=x;a+c-b=y;a+b-c=z\)

=> \(\hept{\begin{cases}a=\frac{y+z}{2}\\b=\frac{z+x}{2}\\c=\frac{x+y}{2}\end{cases}}\)

nên \(M=\frac{1}{2}\left[\frac{4\left(y+z\right)}{x}+\frac{9\left(z+x\right)}{y}+\frac{16\left(x+y\right)}{z}\right]\)

            \(=\frac{1}{2}\left(\frac{4y}{x}+\frac{4z}{x}+\frac{9z}{y}+\frac{9x}{y}+\frac{16x}{z}+\frac{16y}{z}\right)\)

Áp dụng bất đẳng thức cô si ta có 

\(\frac{4y}{x}+\frac{9x}{y}\ge2.\sqrt{\frac{4y.9x}{xy}}=12\)

\(\frac{4z}{x}+\frac{16x}{z}\ge2\sqrt{\frac{4z.16x}{xz}}=2.8=16\)

\(\frac{16y}{z}+\frac{9z}{y}\ge2\sqrt{\frac{16y.9z}{yz}}=2.12=24\)

cộng vào ta có 

\(M\ge\frac{1}{2}\left(12+16+24\right)=26\)

=> \(M\ge26\)

CÁCH KHÁC NÈ MỌI NGƯỜI !!!!!!

\(M+14,5=\frac{4a}{b+c-a}+2+\frac{9b}{a+c-b}+4,5+\frac{16c}{a+b-c}+8\)

=> \(M+14,5=\frac{4a+2\left(b+c-a\right)}{b+c-a}+\frac{9b+4,5\left(a+c-b\right)}{a+c-b}+\frac{16c+8\left(a+b-c\right)}{a+b-c}\)

=> \(M+14,5=\frac{2\left(a+b+c\right)}{b+c-a}+\frac{4,5\left(a+b+c\right)}{a+c-b}+\frac{8\left(a+b+c\right)}{a+b-c}\)

=> \(M+14,5=\left(a+b+c\right)\left(\frac{2}{b+c-a}+\frac{4,5}{a+c-b}+\frac{8}{a+b-c}\right)\)

=> \(M+14,5\ge\frac{\left(a+b+c\right)\left(\sqrt{2}+\sqrt{4,5}+\sqrt{8}\right)^2}{a+b-c+b+c-a+c+a-b}\)     (BĐT CAUCHY - SCHWARZ)

=> \(M+14,5\ge\frac{a+b+c}{a+b+c}.40,5\)

=> \(M+14,5\ge40,5\)

=> \(M\ge40,5-14,5=26\)

VẬY GIÁ TRỊ NHỎ NHẤT CỦA M LÀ 26.

Ban nen cho phan khac chu khong phai phan giai tri

b+c-a > 0 

a + c - b > 0 

a + b - c > 0 

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

=> x + y / 2  = c 

y+z/2 = a 

x+z/2 = b

Khi đó , P = \(\frac{4\frac{\left(y+z\right)}{2}}{x}+\frac{9\frac{x+z}{2}}{y}+\frac{16\frac{x+y}{2}}{z}\)

\(=\frac{1}{2}\left[\frac{4\left(y+z\right)}{x}+\frac{9\left(x+z\right)}{y}+\frac{16\left(x+y\right)}{z}\right]\)

\(=\frac{1}{2}\left[\left(\frac{4y}{x}+\frac{9x}{y}\right)+\left(\frac{4z}{x}+\frac{16x}{z}\right)+\left(\frac{9z}{y}+\frac{16y}{z}\right)\right]\)

Tới đây dễ rồi nha , áp dụng bđt cô - si nha anh 

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:

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

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

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

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

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

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

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

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

\(P\ge2\sqrt{\frac{18xy}{2xy}}+2\sqrt{\frac{16xz}{xz}}+2\sqrt{\frac{72yz}{2yz}}=26\)

Dấu "=" xảy ra khi \(x=\frac{2y}{3}=\frac{z}{2}\)

Đặt \(\hept{\begin{cases}b+c-a=2x\\c+a-b=2y\\a+b-c=2z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=y+z\\b=x+z\\c=x+y\end{cases}}\)

\(\Rightarrow P=\frac{1}{2}.\left(\frac{4\left(y+z\right)}{x}+\frac{9\left(x+z\right)}{y}+\frac{16\left(x+y\right)}{z}\right)\)

\(=\frac{1}{2}.\left(\left(\frac{4y}{x}+\frac{9x}{y}\right)+\left(\frac{4z}{x}+\frac{16x}{z}\right)+\left(\frac{9z}{y}+\frac{16y}{z}\right)\right)\)

\(\ge\frac{1}{2}.\left(2.2.3+2.2.4+2.3.4\right)=26\)

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