\(0\le x,y,z\le1\Rightarrow x^{10}\le x;y^6\le y;z^{2016}\le z;0\le xyz\le1\)

CÓ: \(\left(1-x\right)\left(1-y\right)\left(1-z\right)\ge0\)

=>\(1-xyz+\left(xy+yz+zx\right)-\left(x+y+z\right)\ge0\)

=>\(x+y+z-xy-yz-zx-xyz\le1\)

=>\(x^{10}+y^6+z^{2016}-xy-yz-zx\le1\)

Dấy "=" xảy ra <=> trong 3 số x,y,z có 1 số bằng 0, 2 số bằng 1 hoặc 1 số bằng 1, 2 số bằng 0

không biết :))))

ủa đây là toám lớp 1 hả anh

cauchy phần mẫu @@

Forever_Alone tên là Anh nhưng ko bt họ

\(P=\sqrt{y}\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}=\left(6-\sqrt{x}-\sqrt{z}\right)\left(\sqrt{x}+2\sqrt{z}\right)+3\sqrt{zx}\)

\(P=-x+6\sqrt{x}-2z+12z=-\left(\sqrt{x}-3\right)^2-2\left(\sqrt{z}-3\right)^2+27\le27\)

\(P_{max}=27\) khi \(\left(x;y;z\right)=\left(9;0;9\right)\)

Áp dụng bất đẳng thức Cauchy-Schwarz, ta được:

\(\left(9x^3+3y^2+z\right)\left(\frac{1}{9x}+\frac{1}{3}+z\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow\frac{x}{9x^3+3y^2+z}\le\frac{x\left(\frac{1}{9x}+\frac{1}{3}+z\right)}{\left(x+y+z\right)^2}=\frac{\frac{1}{9}+\frac{x}{3}+zx}{\left(x+y+z\right)^2}\)(1)

Hoàn toàn tương tự, ta có: \(\frac{y}{9y^3+3z^2+x}\le\frac{\frac{1}{9}+\frac{y}{3}+xy}{\left(x+y+z\right)^2}\)(2); \(\frac{z}{9z^3+3x^2+y}\le\frac{\frac{1}{9}+\frac{z}{3}+yz}{\left(x+y+z\right)^2}\)(3)

Cộng theo vế của 3 bất đẳng thức (1), (2), (3), ta được:

\(\frac{x}{9x^3+3y^2+z}+\frac{y}{9y^3+3z^2+x}+\frac{z}{9z^3+3x^2+y}\)\(\le\frac{\frac{1}{9}.3+\frac{x+y+z}{3}+xy+yz+zx}{\left(x+y+z\right)^2}\)

\(\le\frac{\frac{1}{9}.3+\frac{x+y+z}{3}+\frac{\left(x+y+z\right)^2}{3}}{\left(x+y+z\right)^2}=1\)(*)

Mặt khác, có: \(2017\left(xy+yz+zx\right)\le2017.\frac{\left(x+y+z\right)^2}{3}=\frac{2017}{3}\)(**)

Từ (*) và (**) suy ra \(A=\frac{x}{9x^3+3y^2+z}+\frac{y}{9y^3+3z^2+x}+\frac{z}{9z^3+3x^2+y}+2017\left(xy+yz+zx\right)\)

\(\le1+\frac{2017}{3}=\frac{2020}{3}\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)

\(\sqrt{2x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{1}{2}\left(x+y+x+z\right)=\dfrac{1}{2}\left(2x+y+z\right)\)

Tương tự: \(\sqrt{2y+xz}\le\dfrac{1}{2}\left(x+2y+z\right)\) ; \(\sqrt{2z+xy}\le\dfrac{1}{2}\left(x+y+2z\right)\)

Cộng vế:

\(P\le\dfrac{1}{2}\left(4x+4y+4z\right)=4\)

\(P_{max}=4\) khi \(x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

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

\(=\sqrt{3.\left(4+xy+yz+zx\right)}\)

Đã biết x² + y² + z² \(\ge\)xy + yz + zx

=> xy + yz + zx \(\le\dfrac{\left(x+y+z\right)^2}{3}\)

Khi đó \(P\le\sqrt{3\left(4+xy+yz+zx\right)}\le\sqrt{3\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}\)

= 4 

Dấu "=" xảy ra <=> x = 2/3 

\(\sqrt{2x+yz}=\sqrt{\left(x+y+z\right)x+yz}=\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{x+2y+z}{2}\\ \Leftrightarrow P=\sum\sqrt{2x+yz}\le\dfrac{x+2y+z+2x+y+z+x+y+2z}{2}=\dfrac{4\left(x+y+z\right)}{2}=2\cdot2=4\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{2}{3}\)

P = \(1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(Bunyacovski)

\(=\sqrt{3\left[4+\left(xy+yz+zx\right)\right]}\)

\(\le\sqrt{3.\left[4+\dfrac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3.\left(4+\dfrac{4}{3}\right)}\) = 4

Dấu "=" xảy ra <=> x = y = z = 2/3 

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)