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\(A=\sqrt{\frac{x}{2y^2z^2+xyz}}+\sqrt{\frac{y}{2x^2z^2+xyz}}+\sqrt{\frac{z}{2x^2y^2+xyz}}\)
\(A=\sqrt{\frac{x^2}{2xyz.yz+xz.xy}}+\sqrt{\frac{y^2}{2xyz.xz+xy.yz}}+\sqrt{\frac{z^2}{2xyz.xy+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{yz\left(xy+yz+xz\right)+xz.xy}}+\sqrt{\frac{y^2}{xz\left(xy+yz+xz\right)+xy.yz}}+\sqrt{\frac{z^2}{xy\left(xy+yz+xz\right)+xz.yz}}\)
\(A=\sqrt{\frac{x^2}{\left(yz+xy\right)\left(yz+xz\right)}}+\sqrt{\frac{y^2}{\left(xz+xy\right)\left(xz+yz\right)}}+\sqrt{\frac{z^2}{\left(xy+yz\right)\left(xy+xz\right)}}\)
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 \(x=y=z=1\)
\(\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2z^2+3}+\frac{1}{z^2+2x^2+3}\)
= \(\frac{1}{x^2+y^2+y^2+1+2}+\frac{1}{y^2+z^2+z^2+1+2}+\frac{1}{z^2+x^2+x^2+1+2}\)
\(\le\frac{1}{2xy+2y+2}+\frac{1}{2yz+2z+2}+\frac{1}{2zx+2x+2}\)
= \(\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)
= \(\frac{1}{2}\left(\frac{zx}{xyzx+yzx+zx}+\frac{x}{yzx+zx+x}+\frac{1}{zx+x+1}\right)\)
= \(\frac{1}{2}\left(\frac{zx}{x+1+zx}+\frac{x}{1+zx+x}+\frac{1}{zx+x+1}\right)\)
= 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
a) + \(x^3+y^3+1=\left(x+y\right)\left(x^2-xy+y^2\right)+1\ge\left(x+y\right)\left(2xy-xy\right)+xyz=xy\left(x+y+z\right)\)
Dấu "=" \(\Leftrightarrow x=y\)
+ Tương tự : \(y^3+z^3+1\ge yz\left(x+y+z\right)\). Dấu "=" \(\Leftrightarrow y=z\)
\(z^3+x^3+1\ge xz\left(x+y+z\right)\). Dấu "=" \(\Leftrightarrow x=z\)
Do đó: \(A\le\frac{1}{xy\left(x+y+z\right)}+\frac{1}{yz\left(x+y+z\right)}+\frac{1}{zx\left(x+y+z\right)}=\frac{x+y+z}{xyz\left(x+y+z\right)}=1\)
Dấu "=" \(\Leftrightarrow x=y=z=1\)
b) Bn đã từng hỏi và cũng là mk trả lời 
Ta có:
\(P=\frac{1}{x^2+2y+3}+\frac{1}{y^2+2z+3}+\frac{1}{z^2+2x+3}\)
\(\le\frac{1}{2}\left(\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\right)\)
Đặt \(x=a^3;y=b^3;z=c^3\)
\(\Rightarrow abc=1\)
Từ đây ta có:
\(2P\le\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}+\frac{1}{c^3+a^3+1}\)
\(=\frac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)+1}+\frac{1}{\left(b+c\right)\left(b^2-bc+c^2\right)+1}+\frac{1}{\left(c+a\right)\left(c^2-ca+a^2\right)+1}\)
\(\le\frac{1}{\left(a+b\right)ab+1}+\frac{1}{\left(b+c\right)bc+1}+\frac{1}{\left(c+a\right)ac+1}\)
\(=\frac{c}{a+b+c}+\frac{a}{a+b+c}+\frac{b}{a+b+c}=1\)
Vậy \(P\le\frac{1}{2}\)
mình nghĩ đề sai vì hôm kia đến h nghĩ mãi không ra D: