x/2=2y/3=3z/4=k,x=2k,y=3k:2,z=4k:3 ta có 2k-3k:2=15=1k:2=15;k=15.2=30;x=2.30=60;y=3.30:2=45,z=4.30:=40_x-y-z=-25

suy ra x-1/2+1=y-2/3+1=z-3/4+1 suy ra x+1/2=y+1/3=z+1/4 = K Ta có x=2K-1;y=3K-1;z=4k-1 mà x-2y+3z =-10

Chứng minh một số bất đẳng thức phụ:

1. \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\ge3\)

2. \(2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\text{ (vừa chứng minh ở trên)}\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\)

3. \(x^2+y^2+z^2\ge xy+yz+zx\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge3\left(xy+y+zx\right)\)

\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow x+y+z\ge\sqrt{3\left(xy+yz+zx\right)}\ge\sqrt{3.3}=3\)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{x^4}{y+3z}+\frac{y^4}{z+3x}+\frac{z^4}{x+3y}\ge\frac{\left(x^2+y^2+z^2\right)^2}{y+3z+z+3x+x+3y}=\frac{\left(x^2+y^2+z^2\right)\left(x^2+y^2+z^2\right)}{4\left(x+y+z\right)}\)

\(\ge\frac{3.\frac{1}{3}\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\frac{x+y+z}{4}\ge\frac{3}{4}\)

Dấu "=" xảy ra khi và chỉ khi x = y = z = 1.

C2: Áp dụng Co6si:

\(\frac{x^4}{y+3z}+\frac{y+3z}{16}+\frac{1}{4}+\frac{1}{4}\ge4\sqrt[4]{\frac{x^4}{y+3z}.\frac{y+3z}{16}.\frac{1}{4}.\frac{1}{4}}=x\)

\(\Rightarrow\frac{x^4}{y+3z}\ge x-\frac{y+3z}{16}-\frac{1}{2}\)

Tương tự \(\frac{y^4}{z+3x}\ge y-\frac{z+3x}{16}-\frac{1}{2};\frac{z^4}{x+3y}\ge z-\frac{x+3y}{16}-\frac{1}{2}\)

\(\Rightarrow\frac{x^4}{y+3z}+\frac{y^4}{z+3x}+\frac{z^4}{x+3y}\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{2}\ge\frac{3}{4}.3-\frac{3}{2}=\frac{3}{4}\)

(\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)\ge xy+yz+zx+2\left(xy+yz+zx\right)\)

\(=3\left(xy+yz+zy\right)\ge9\)

\(\Rightarrow x+y+z\ge3\))

Dấu "=" xảy ra khi x = y = z = 1.

e, Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)

\(\Rightarrow\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{\left(x-2y+3z\right)+\left(-1+4-9\right)}{8}=\frac{14-6}{8}=1\)

Do đó: \(\frac{x-1}{2}=1\Rightarrow x=2.1+1=3\)

\(\frac{2y-4}{6}=1\Rightarrow y=\frac{6.1+4}{2}=5\)

\(\frac{3z-9}{12}=1\Rightarrow z=\frac{12.1+9}{3}=7\)

Vậy x=3; y=5; z=7

h, Ta có: \(\frac{x}{2}=\frac{y}{3}=\left(\frac{x}{2}\right)^2=\left(\frac{y}{3}\right)^2=\frac{x^2}{4}=\frac{y^2}{9}=\frac{x.y}{2.3}=\frac{54}{6}=9\)

Do đó: \(\frac{x^2}{4}=9\Rightarrow x^2=4.9=36\Rightarrow x=6;x=-6\)

\(\frac{y^2}{9}=9\Rightarrow y^2=9.9=81\Rightarrow y=9;y=-9\)

Áp dụng AM-GM ta có \(\frac{1^2}{x}+\frac{1^2}{x}+\frac{1^2}{y}+\frac{1^2}{z}\ge\frac{\left(1+1+1+1\right)^2}{2x+y+z}\)

hay \(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{16}{2x+y+z}\)

Tương tự : \(\frac{2}{y}+\frac{1}{x}+\frac{1}{z}\ge\frac{16}{2y+x+z}\) ; \(\frac{2}{z}+\frac{1}{x}+\frac{1}{y}\ge\frac{16}{2z+x+y}\)

Cộng theo vế : \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge16\left(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\right)\)

\(\Leftrightarrow\)\(16\left(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\right)\le16\)

\(\Leftrightarrow\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\le1\)