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\(xy+yz+zx\le3xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3\)
\(P=\frac{1}{\sqrt{x^2+y^2+x^2+xy}}+\frac{1}{\sqrt{y^2+z^2+y^2+yz}}+\frac{1}{\sqrt{z^2+x^2+z^2+zx}}\)
\(P\le\frac{1}{\sqrt{x^2+3xy}}+\frac{1}{\sqrt{y^2+3yz}}+\frac{1}{\sqrt{z^2+3zx}}=\frac{4}{2\sqrt{4x\left(x+3y\right)}}+\frac{4}{2\sqrt{4y\left(y+3z\right)}}+\frac{1}{2\sqrt{4z\left(z+3x\right)}}\)
\(P\le4\left(\frac{1}{4x+x+3y}+\frac{1}{4y+y+3z}+\frac{1}{4z+z+3x}\right)=4\left(\frac{1}{5x+3y}+\frac{1}{5y+3z}+\frac{1}{5z+3x}\right)\)
\(P\le\frac{4}{64}\left(\frac{5}{x}+\frac{3}{y}+\frac{5}{y}+\frac{3}{z}+\frac{5}{z}+\frac{3}{x}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{3}{2}\)
\(P_{max}=\frac{3}{2}\) khi \(x=y=z=1\)
Từ: \(xy+yz+xz=xyz\) <=> \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(A=\frac{1}{x+2y+3z}+\frac{1}{2x+3y+z}+\frac{1}{2x+y+2z}\)
Áp dụng bđt: \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) (tự cm đúng)
Ta có: \(\frac{1}{x+2y+3z}=\frac{1}{x+z+2y+2z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{2y+2z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{2y}+\frac{3}{2z}\right)\) (1)
CMTT: \(\frac{1}{2x+3y+z}\le\frac{1}{16}\left(\frac{1}{2x}+\frac{1}{z}+\frac{3}{2y}\right)\) (2)
\(\frac{1}{3x+y+2z}\le\frac{1}{16}\left(\frac{3}{2x}+\frac{1}{y}+\frac{1}{2z}\right)\)(3)
Từ (1); (2) và (3) cộng vế theo vế
\(A\le\frac{1}{16}\left(\frac{3}{2z}+\frac{1}{x}+\frac{1}{2y}+\frac{3}{2y}+\frac{1}{z}+\frac{1}{2x}+\frac{3}{2z}+\frac{1}{y}+\frac{1}{2z}\right)\)
\(A\le\frac{3}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3}{16}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y+2z\\z=2x+y\\y=x+2z\end{cases}}\) <=> x = y = z = 0
mà x;y;z > 0 => Dấu "=" ko xảy ra
=> A < 3/16
Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).
\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).
\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).
Ta có:
\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)
\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).
Ta có:
\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).
\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).
\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).
\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).
\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).
Chứng minh tương tự, ta được:
\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).
Chứng minh tương tự, ta được:
\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).
Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:
\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).
\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).
\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)
\(\left(4\right)\).
Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).
\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)
(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).
\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).
Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:
\(P\ge\frac{\sqrt{5}}{3}\).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).
Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).
\(A=\frac{xy+2y+1}{xy+x+y+1}+\frac{yz+2z+1}{yz+y+z+1}+\frac{zx+2x+1}{zx+z+x+1}\)
\(=\frac{y\left(x+1\right)+y+1}{\left(x+1\right)\left(y+1\right)}+\frac{z\left(y+1\right)+z+1}{\left(y+1\right)\left(z+1\right)}+\frac{x\left(z+1\right)+x+1}{\left(z+1\right)\left(x+1\right)}\)
\(=\frac{y}{y+1}+\frac{1}{x+1}+\frac{z}{z+1}+\frac{1}{y+1}+\frac{x}{x+1}+\frac{1}{z+1}\)
\(=\frac{y+1}{y+1}+\frac{z+1}{z+1}+\frac{x+1}{x+1}=3\)
ta có: \(\frac{\sqrt{2x^2+y^2}}{xy}=\sqrt{\frac{2}{y^2}+\frac{1}{x^2}}\)
Áp dụng BĐT bunyakovsky:\(\left(2+1\right)\left(\frac{2}{y^2}+\frac{1}{x^2}\right)\ge\left(\frac{2}{y}+\frac{1}{x}\right)^2\)
\(\Rightarrow\frac{2}{y^2}+\frac{1}{x^2}\ge\frac{1}{3}\left(\frac{2}{y}+\frac{1}{x}\right)^2\).....bla bla
b) Ta có \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{y+z+z+x+x+y}\)(BĐT Schwarz)
\(=\frac{x+y+z}{2}=\frac{2}{2}=1\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^2}{y+z}=\frac{y^2}{z+x}=\frac{z^2}{x+y}\\x+y+z=2\end{cases}}\Leftrightarrow x=y=z=\frac{2}{3}\)
a) Có \(P=1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(BĐT Bunyakovsky)
\(=\sqrt{3.\left[2\left(x+y+z\right)+xy+yz+zx\right]}\)
\(\le\sqrt{3\left[4+\frac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3\left(4+\frac{4}{3}\right)}=4\)
Dấu "=" xảy ra <=> x = y = z = 2/3
t acó \(xy+yz+zx=xyz\Rightarrow\) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
đặt biểu thức =A
Áp dụng bất dẳng thức Svác sơta có \(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{z}\ge\frac{36}{x+2y+3z}\)
tương tự , ta có
\(\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{x}+\frac{1}{x}+\frac{1}{x}\ge\frac{36}{y+2x+3z}\)
\(\frac{1}{z}+\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{y}\ge\frac{36}{z+2x+3y}\)
cộng từng vế của 3 bđt cùng chiều ta có \(36A\ge6\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=6\)
\(\Rightarrow A\ge\frac{1}{6}\)
dấu = xảy ra <=> x=y=z=3
=99/10