Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm:

\(\Rightarrow\left\{\begin{matrix}3yzt\le y^3+z^3+t^3\\3xtz\le x^3+t^3+z^3\\3xyt\le x^3+y^3+t^3\\3xyz\le x^3+y^3+z^3\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x^3+3yzt\le x^3+y^3+z^3+t^3\\y^3+3xtz\le x^3+y^3+z^3+t^3\\z^3+3xyt\le x^3+y^3+z^3+t^3\\t^3+3xyz\le x^3+y^3+z^3+t^3\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\frac{x^3}{x^3+3yzt}\ge\frac{x^3}{x^3+y^3+z^3+t^3}\\\frac{y^3}{y^3+3xtz}\ge\frac{y^3}{x^3+y^3+z^3+t^3}\\\frac{z^3}{z^3+3xyt}\ge\frac{z^3}{x^3+y^3+z^3+t^3}\\\frac{t^3}{t^3+3xyz}\ge\frac{t^3}{x^3+y^3+z^3+t^3}\end{matrix}\right.\)

\(\Rightarrow\frac{x^3}{x^3+3yzt}+\frac{y^3}{y^3+3xtz}+\frac{z^3}{z^3+3xyt}+\frac{t^3}{t^3+3xyz}\ge\frac{x^3+y^3+z^3+t^3}{x^3+y^3+z^3+t^3}\)

\(\Rightarrow\frac{x^3}{x^3+3yzt}+\frac{y^3}{y^3+3xtz}+\frac{z^3}{z^3+3xyt}+\frac{t^3}{t^3+3xyz}\ge1\) ( đpcm )

Câu trả lời cần bổ sung : dấu bằng xảy ra khi và chỉ khi x = y = z = t > 0

đề có cho điều kiện ko

\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)

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

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

\(\Rightarrow A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)=\left(x+y+z\right)-\frac{xy+yz+zx}{2}\ge3-\frac{\frac{9}{3}}{2}=\frac{3}{2}\)

Dau '=' xay ra khi \(x=y=z=1\)

Vay \(A_{min}=\frac{3}{2}\)khi \(x=y=z=1\)

Ta sẽ chứng minh: \(\frac{a^3}{a^2+ab+b^2}\ge\frac{2a-b}{3}\) với a;b dương

Thật vậy, BĐT tương đương:

\(3a^3\ge\left(2a-b\right)\left(a^2+ab+b^2\right)\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)

Áp dụng: \(\Rightarrow S\ge\frac{2x-y}{3}+\frac{2y-z}{3}+\frac{2z-x}{3}=\frac{x+y+z}{3}=3\)

\(S_{min}=3\) khi \(x=y=z=3\)

\(\Sigma\frac{x^3}{y^2}=\Sigma\frac{x}{y^2}\left(x-y\right)^2+\frac{\Sigma z\left(x^3-yz^2\right)^2}{xyz\left(x+y+z\right)}+\Sigma\frac{x^2}{y}\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\)

\(VT-VP=\Sigma\frac{\left(x+y\right)\left(x-y\right)^2}{y^2}\ge0\)

\(x\left(x-z\right)+y\left(y-z\right)=0\)\(\Leftrightarrow\)\(x^2+y^2=z\left(x+y\right)\)

\(\frac{x^3}{z^2+x^2}=x-\frac{z^2x}{z^2+x^2}\ge x-\frac{z^2x}{2zx}=x-\frac{z}{2}\)

\(\frac{y^3}{y^2+z^2}=y-\frac{yz^2}{y^2+z^2}\ge y-\frac{yz^2}{2yz}=y-\frac{z}{2}\)

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

Cộng lại ra minP=4, dấu "=" xảy ra khi \(x=y=z=1\)

Cauchy ngược dấu:v

\(A\ge x\left(1-\frac{y}{2}\right)+y\left(1-\frac{z}{2}\right)+z\left(1-\frac{x}{2}\right)\)

\(=x+y+z-\frac{xy+yz+zx}{2}\ge3-\frac{\left(x+y+z\right)^2}{6}=\frac{3}{2}\)

Đẳng thức xảy ra khi x = y = z = 1.

P/s: Ko chắc~