\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)

\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)

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

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

Áp dụng bđt \(\sqrt{ab}\le\frac{a+b}{2}\) ta có:

\(2A\le\frac{x}{yz+xy}+\frac{x}{yz+xz}+\frac{y}{xz+xy}+\frac{y}{xz+yz}+\frac{z}{xy+yz}+\frac{z}{xy+xz}\)

\(=\frac{x+z}{yz+xy}+\frac{x+y}{yz+xz}+\frac{y+z}{xz+xy}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Mà: \(xy+yz+xz=2xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Rightarrow2A\le2\Rightarrow A\le1."="\Leftrightarrow a=b=c=\frac{3}{2}\)

\(\frac{P}{\sqrt{6}}=\sum\frac{1}{\sqrt{6}}.\frac{1}{\sqrt{2x^2+y^2+3}}\le\frac{1}{2}\sum\left(\frac{1}{6}+\frac{1}{2x^2+y^2+3}\right)\)

\(\frac{P}{\sqrt{6}}\le\frac{1}{4}+\frac{1}{2}\sum\frac{1}{2\left(x^2+1\right)+\left(y^2+1\right)}\le\frac{1}{4}+\frac{1}{2}\sum\frac{1}{4x+2y}\)

\(\frac{P}{\sqrt{6}}\le\frac{1}{4}+\frac{1}{4}\sum\frac{1}{x+x+y}\le\frac{1}{4}+\frac{1}{36}\left(\frac{2}{x}+\frac{1}{y}+\frac{2}{y}+\frac{1}{z}+\frac{2}{z}+\frac{1}{x}\right)\)

\(\frac{P}{\sqrt{6}}\le\frac{1}{4}+\frac{1}{12}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\)

\(\Rightarrow P\le\frac{\sqrt{6}}{2}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

uy bạn giỏi thế lớp 7 học toán 8 rồi af gh3 z

Đặt \(^{\hept{\begin{cases}x=a^2\\y=b^2\\z=c^2\end{cases}}\Rightarrow abc=1}\)

\(\Rightarrow P=\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\)

ÁP DỤNG BĐT AM-GM : 

\(a^2+b^2\ge2ab\)

\(b^2+1\ge2b\)

\(\Rightarrow a^2+2b^2+3\ge2\left(ab+b+1\right)\)

\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2}.\frac{1}{ab+b+1}\)

Tương tự \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}.\frac{1}{bc+c+1}\)

               \(\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ac+a+1}\)

Cộng từng vế các bđt trên ta được

\(P\le\frac{1}{2}\)

Dấu "=" xảy ra khi x=y=z=1

\(\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}\)

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

\(\le\frac{1}{2xy+2y+2}+\frac{1}{2yz+2z+2}+\frac{1}{2zx+2x+2}\)

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

= \(\frac{1}{2}\left(\frac{zx}{xyzx+yzx+zx}+\frac{x}{yzx+zx+x}+\frac{1}{zx+x+1}\right)\)

= \(\frac{1}{2}\left(\frac{zx}{x+1+zx}+\frac{x}{1+zx+x}+\frac{1}{zx+x+1}\right)\)

= 1/2

Dấu "=" xảy ra <=> x = y =z =1 

Áp dụng BĐT AM-GM ta có:\(\hept{\begin{cases}x^2+y^2\ge2xy\\y^2+1\ge2y\end{cases}\Rightarrow\frac{1}{x^2+2y^2+3}\le\frac{1}{2xy+2y+2}}\)

Tương tự ta cũng có

\(\frac{1}{y^2+2x^2+3}\le\frac{1}{2yz+2z+2};\frac{1}{z^2+2x^2+3}\le\frac{1}{2xz+2x+2}\)

Do đó ta có:\(VT\le\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)

Mặt khác, do xyz=1 nên ta có:

\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}=\frac{1}{xy+y+1}+\frac{y}{xy+y+1}+\frac{xy}{xy+y+1}\)

\(=\frac{xy+y+1}{xy+y+1}=1\)

\(\Rightarrow VT\le\frac{1}{2}\). Dấu "=" xảy ra <=> x=y=z=1