\(P^2=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2.\left(\frac{xy.yz}{zx}+\frac{yz.zx}{xy}+\frac{zx.xy}{zy}\right)\)

\(=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2.2016\)

Áp dụng BĐT Cauchy:\(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}\ge2\sqrt{\frac{x^2y^2}{z^2}.\frac{y^2z^2}{x^2}}=2y^2\)

\(\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge2\sqrt{\frac{y^2z^2}{x^2}.\frac{z^2x^2}{y^2}}=2z^2\)

\(\frac{z^2x^2}{y^2}+\frac{x^2y^2}{z^2}\ge2\sqrt{\frac{x^2z^2}{y^2}.\frac{x^2y^2}{z^2}}=2x^2\)

Cộng theo vế ta được:\(2\left(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\right)\ge2x^2+2y^2+2z^2=2.2016\)

\(\Rightarrow\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge2016\)

\(\Rightarrow P^2\ge2016+2016.2=6048\Rightarrow P\ge\sqrt{6048}=12\sqrt{42}\)

Nên GTNN của P là \(12\sqrt{42}\) đạt được khi \(x=y=z=\sqrt{\frac{2016}{3}}=4\sqrt{42}\)

\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)

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

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

\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Ta có: \(xy+yz+zx=xyz\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)ta có: \(a,b,c>0;a+b+c=1\)do đó 0<a,b,c<1

\(P=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+6\left(ab+bc+ca\right)\)

\(=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+2\left(a+b+c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)

\(=\left(\frac{b^2}{a}-2b+a\right)+\left(\frac{c^2}{b}-2c+b\right)+\left(\frac{a^2}{c}-2a+c\right)-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)

\(=\frac{\left(a-b\right)^2}{a}+\frac{\left(b-c\right)^2}{b}+\frac{\left(c-a\right)^2}{c}-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)

\(=\frac{\left(1-a\right)\left(a-b\right)^2}{a}+\frac{\left(1-b\right)\left(b-c\right)^2}{b}+\frac{\left(1-c\right)\left(c-a\right)^2}{c}+3\ge3\)

Vậy GTNN của P=3

min hay max bạn

Mk nghĩ là x³,y³,z³.

Áp dụng BĐT AM-GM:

\(\Sigma_{cyc}\left(\frac{x^2}{\sqrt{x^3+8}}\right)=\Sigma_{cyc}\left(\frac{x^2}{\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}}\right)\)\(\ge2\Sigma_{cyc}\left(\frac{x^2}{x^2-x+6}\right)\)

Áp dụng BĐT Cauchy-Schwart:

\(2\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)\(=\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-2\left(xy+yz+zx\right)-\left(x+y+z\right)+18}\)\(\ge\frac{2\left(x+y+z\right)^2}{\left(x+y+z\right)^2-2\left(x+y+z\right)-\left(x+y+z\right)+18}\)

gt\(\Leftrightarrow3\left(x+y+z\right)\le3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2\)

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

\(\Rightarrow\left[{}\begin{matrix}x+y+z\le0\\x+y+z\ge3\end{matrix}\right.\)

Đặt t=x+y+z\(\left(t\ge3\right)\)

Cần c/m:\(\frac{2t^2}{t^2-3t+18}\ge1\)

Có :\(t^2-3t+18>0\)

\(\Rightarrow2t^2\ge t^2-3t+18\)

\(\Leftrightarrow t^2+3t-18\ge3^2+3.3-18=0\)(Đúng)

Vậy min =1

Dấu = xra khi x=y=z=1.

#Walker

Kiểm tra giùm em đúng ko ạ Akai Haruma

\(\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}=\frac{x}{x\left(x+y+z\right)+yz}+\frac{y}{y\left(x+y+z\right)+zx}+\frac{z}{z\left(x+y+z\right)+xy}\)

\(=\text{Σ}\frac{x}{\left(x+y\right)\left(x+z\right)}=\frac{2\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\)(1)

+) CM bổ đề (cái này khá hữu dụng): \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}\cdot3\sqrt[3]{x^2y^2z^2}=9xyz\Leftrightarrow\frac{1}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\ge xyz\)

Có \(\left(x+y\right)\left(y+z\right)\left(x+z\right)=\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)

Thay vào (1)-> DPCM

Dấu = xảy ra khi x=y=z=1/3

Thx HD film