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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\geq \frac{(x+y+z)^2}{x+y+y+z+z+x}\)
\(\Leftrightarrow A\geq \frac{x+y+z}{2}\)
Áp dụng BĐT AM-GM:
\(\left\{\begin{matrix} x+y\geq 2\sqrt{xy}\\ y+z\geq 2\sqrt{yz}\\ z+x\geq 2\sqrt{zx}\end{matrix}\right.\)
\(\Rightarrow 2(x+y+z)\geq 2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})=2\)
\(\Rightarrow x+y+z\geq 1\)
Do đó: \(A\geq \frac{x+y+z}{2}\geq \frac{1}{2}\)
Vậy \(A_{\min}=\frac{1}{2}\)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
\(Q=\dfrac{xyz}{z^3\left(x+y\right)}+\dfrac{xyz}{x^3\left(y+z\right)}+\dfrac{xyz}{y^3\left(x+z\right)}\)
\(=\dfrac{1}{z^3\left(x+y\right)}+\dfrac{1}{y^3\left(x+z\right)}+\dfrac{1}{x^3\left(y+z\right)}\) (vì xyz = 1)
\(=\dfrac{\left(\dfrac{1}{z}\right)^2}{z\left(x+y\right)}+\dfrac{\left(\dfrac{1}{y}\right)^2}{y\left(x+z\right)}+\dfrac{\left(\dfrac{1}{x}\right)^2}{x\left(y+z\right)}\)
Áp dụng BĐT cauchy schwarz với x,y,z > 0 ta có:
\(Q\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(xy+yz+xz\right)}=\dfrac{\left(xy+yz+xz\right)^2}{2\left(xy+yz+xz\right)}=\dfrac{xy+yz+xz}{2}\)Mặt khác theo BĐT cauchy với x;y;z>0 thì
\(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}=3\)
Vậy MinQ = \(\dfrac{3}{2}\Leftrightarrow x=y=z=1\)
Cái bài này bình thường :v
Đặt \(A=\dfrac{x^3}{y^3+8}+\dfrac{y^3}{z^3+8}+\dfrac{z^3}{x^3+8}\)
\(BDT\Leftrightarrow\dfrac{x^3}{y^3+8}+\dfrac{y^3}{z^3+8}+\dfrac{z^3}{x^3+8}-\dfrac{2}{27}\left(xy+yz+xz\right)\ge\dfrac{1}{9}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{x^3}{y^3+8}+\dfrac{y+2}{27}+\dfrac{y^2-2y+4}{27}\)
\(\ge3\sqrt[3]{\dfrac{x^3}{y^3+8}\cdot\dfrac{y+2}{27}\cdot\dfrac{y^2-2y+4}{27}}=\dfrac{x}{3}\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\dfrac{y^3}{z^3+8}+\dfrac{z+2}{27}+\dfrac{z^2-2z+4}{27}\ge\dfrac{y}{3};\dfrac{z^3}{x^3+8}+\dfrac{x+2}{27}+\dfrac{x^2-2x+4}{27}\ge\dfrac{z}{3}\)
Cộng theo vế 3 BĐT trên ta có:
\(A+\dfrac{x+y+z+6}{27}+\dfrac{x^2+y^2+z^2-2\left(x+y+z\right)+12}{27}\ge\dfrac{x+y+z}{3}\)
\(\Leftrightarrow A+\dfrac{9}{27}+\dfrac{\dfrac{\left(x+y+z\right)^2}{3}+6}{27}\ge1\)\(\Leftrightarrow A\ge\dfrac{1}{3}\)
Cần chứng minh \(VT=A-\dfrac{2}{27}\left(xy+yz+xz\right)\ge\dfrac{1}{9}=VP\)
\(\Leftrightarrow VT=\dfrac{1}{3}-\dfrac{2\cdot\dfrac{\left(x+y+z\right)^2}{3}}{27}=\dfrac{1}{9}=VP\) (đúng)
Xảy ra khi \(x=y=z=1\)
P/s:Trình bày hơi khó hiểu, thông cảm :v
Ta có : Áp dụng BĐT Cauchy ba số ở mẫu ta được
\(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\dfrac{x}{\dfrac{y+z+1}{3}}+\dfrac{y}{\dfrac{x+z+1}{3}}+\dfrac{z}{\dfrac{x+y+1}{3}}=\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\)Thấy: \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}\left(?!\right)\)
Ta phải chứng minh:
\(\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{3}\)
\(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
Mà \(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}=\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\)
Theo C.B.S
\(\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Phải chứng minh
\(\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
\(\Leftrightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
Ta có : \(xy+yz+xz\le x^2+y^2+z^2=3\)
Theo C.B.S : \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le9\)
\(\Rightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
=> ĐPCM
\(A=\dfrac{2x^2}{2x+2yz}+\dfrac{2y^2}{2y+2zx}+\dfrac{2z^2}{2z+2xy}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)
\(A\ge\dfrac{2x^2}{x^2+1+y^2+z^2}+\dfrac{2y^2}{y^2+1+z^2+x^2}+\dfrac{2z^2}{z^2+1+x^2+y^2}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)
\(A\ge\dfrac{2\left(x^2+y^2+z^2\right)}{x^2+y^2+z^2+1}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)
Đặt \(x^2+y^2+z^2=a>0\)
\(\Rightarrow A\ge\dfrac{2a}{a+1}+\dfrac{9}{8a}=\dfrac{2a}{a+1}+\dfrac{9}{8a}-\dfrac{15}{8}+\dfrac{15}{8}\)
\(\Rightarrow A\ge\dfrac{\left(a-3\right)^2}{8a\left(a+1\right)}+\dfrac{15}{8}\ge\dfrac{15}{8}\)
\(A_{min}=\dfrac{15}{8}\) khi \(a=3\) hay \(x=y=z=1\)
Chỉ em phương pháp múa cột trong tính nguyên hàm với ạ