Cái đề thiếu x, y, z dương bạn nhé

Với mọi x, y, z > 0 ta luôn có

\(x^3+y^3\ge x^2y+xy^2\)    (1)

\(\Leftrightarrow\left(x^3-x^2y\right)+\left(y^3-xy^2\right)\ge0\)

\(\Leftrightarrow x^2\left(x-y\right)+y^2\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2-y^2\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\)   (luôn đúng)

Tương tự  \(y^3+z^3\ge y^2z+yz^2\)   (2)  và   \(z^3+x^3\ge z^2x+zx^2\)   (3)

Cộng (1), (2), (3) vế theo vế ta được  \(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2\le2\left(x^3+y^3+z^3\right)\)

Theo BĐT Cauchy-Schwarz, ta có

\(VT=\frac{x^6}{x^3+x^2y+xy^2}+\frac{y^6}{y^3+y^2z+yz^2}+\frac{z^6}{z^3+z^2x+zx^2}\)

\(\ge\frac{\left(x^3+y^3+z^3\right)^2}{\left(x^3+y^3+z^3\right)+\left(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2\right)}\ge\frac{\left(x^3+y^3+z^3\right)^2}{\left(x^3+y^3+z^3\right)+2\left(x^3+y^3+z^3\right)}\)

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

Đẳng thức xảy ra \(\Leftrightarrow x=y=z\)

x, y, z là số thực anh ơi

\(=\frac{y^2\left(x-y\right)-z^2\left(x-y\right)}{x^2z-x^2y+xy^2-y^2z+yz^2-xz^2}\)

\(\frac{y^2\left(x-y\right)-z^2\left(x-y\right)}{z\left(x^2-y^2\right)-xy\left(x-y\right)-z^2\left(x-y\right)}\)

\(\frac{\left(x-y\right)\left(y^2-z^2\right)}{\left(x-y\right)\left(xz-yz-xy-z^2\right)}\)

tutuwjj làm típ

\(\frac{y}{z}=\frac{14}{3}\Rightarrow\frac{1}{y}=\frac{3}{14.z}\Rightarrow\frac{x}{y}=\frac{3x}{14.z}=\frac{4}{7}\Rightarrow\frac{x}{z}=\frac{4.14}{7.3}=\frac{8}{3}\)

\(\frac{x+y}{z}=\frac{x}{z}+\frac{y}{z}=\frac{8}{3}+\frac{14}{3}=\frac{22}{3}\)

bằng 22/3

\(\left(\frac{x}{z+y}+\frac{y}{z+x}+\frac{z}{x+y}\right)\left(x+y+z\right)=1\\ \)

Nhân phân phối ra

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

\(\left(\frac{x^2}{z+y}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)=0\)

Chờ các bạn lâu quá nên mình giải luôn: (x+y+z)\(\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\)) = \(\frac{x^2}{y+z}+\frac{xy}{x+z}+\frac{xz}{x+y}+\frac{xy}{y+z}+\frac{y^2}{x+z}+\frac{yz}{x+y}+\frac{xz}{y+z}+\frac{yz}{x+z}+\frac{z^2}{x+y}=1\)

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

Do đó: \(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0\)

bạn đưa về 1 ẩn rồi giải nhen :

a) \(\frac{x}{y}=\frac{2}{3}\Rightarrow y=\frac{3x}{2}\)

Ta có : \(x.y=54\Leftrightarrow x.\frac{3x}{2}=54\)

\(\Rightarrow3x^2=108\)

\(\Rightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)

ĐK : x; y; z > 0

Ta có : \(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)\)

\(\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}+2\sqrt{\frac{z}{x}.\frac{x}{z}}+2\sqrt{\frac{y}{z}.\frac{z}{y}}=2+2+2=6\) (BĐT AM - GM) (ĐFCM)

(Dấu "=" xảy ra <=> \(x=y=z\))