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Với 2 số dương bất kì: ( 1 )
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)Vì x và y dương nên \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\forall x;y\)
Áp dụng ( 1 ): \(\frac{4}{2x+y+z}=\frac{4}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{x+y}+\frac{1}{x+z}\)
Mà: \(\frac{1}{x+y}+\frac{1}{x+z}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)=\frac{1}{4}\)\(=\frac{1}{4}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Nên: \(\frac{1}{2x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có: \(\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)
Và \(\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Cộng vế với vế các bất đẳng thức kết hợp với điều kiện \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\) nên ta có đpcm
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
BĐT tương đương:
\(\frac{1}{z\left(1+\frac{1}{x}\right)}+\frac{1}{x\left(1+\frac{1}{y}\right)}+\frac{1}{y\left(1+\frac{1}{z}\right)}\ge2\)
Từ giả thiết:
\(xy+yz+zx+2xyz=1\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+2=\frac{1}{xyz}\)
Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow a+b+c+2=abc\)
\(\Rightarrow a+b+c+2\le\frac{1}{27}\left(a+b+c\right)^3\)
\(\Leftrightarrow\left(a+b+c\right)^3-27\left(a+b+c\right)-54\ge0\)
\(\Leftrightarrow\left(a+b+c-6\right)\left(a+b+c+3\right)^2\ge0\)
\(\Leftrightarrow a+b+c\ge6\)
BĐT trở thành: \(\frac{c}{1+a}+\frac{a}{1+b}+\frac{b}{1+c}\ge2\)
Thật vậy, ta có:
\(VT=\frac{a^2}{a+ab}+\frac{b^2}{b+bc}+\frac{c^2}{c+ca}\ge\frac{\left(a+b+c\right)^2}{a+b+c+ab+bc+ca}\ge\frac{3\left(a+b+c\right)^2}{3\left(a+b+c\right)+\left(a+b+c\right)^2}\)
\(VT\ge\frac{3\left(a+b+c\right)}{3+a+b+c}=\frac{2\left(a+b+c\right)+a+b+c}{a+b+c+3}\ge\frac{2\left(a+b+c\right)+6}{a+b+c+3}=2\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=2\) hay \(x=y=z=\frac{1}{2}\)
\(VT=6\left(x^2+y^2+z^2\right)+10\left(xy+yz+xz\right)+2\left(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\right)\)
\(=6\left(x+y+z\right)^2-2\left(xy+yz+xz\right)+2\frac{9}{2x+y+z+x+2y+z+x+y+2z}\)
\(\ge6\left(x+y+z\right)^2-2\frac{\left(x+y+z\right)^2}{3}+2\frac{9}{4\left(x+y+z\right)}\)
\(=\: 6\cdot\left(\frac{3}{4}\right)^2-2\cdot\frac{\left(\frac{3}{4}\right)^2}{3}+2\cdot\frac{9}{4\cdot\frac{3}{4}}=9\)
+) Áp dụng BĐT Cô - si cho 4 số dương x; x; y; z ta có:
\(x+x+y+z\ge4\sqrt[4]{x.x.y.z}\)
=> 2x + y + z \(\ge4\sqrt[4]{x.x.y.z}\) (1)
Với 4 số dương \(\frac{1}{x};\frac{1}{x};\frac{1}{y};\frac{1}{z}\) ta có: \(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge4.\sqrt[4]{\frac{1}{x}.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}}\) (2)
Từ (1)(2) => \(\left(2x+y+z\right)\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4.\sqrt[4]{x.x.y.z}4.\sqrt[4]{\frac{1}{x}.\frac{1}{x}.\frac{1}{y}.\frac{1}{z}}=16\)
=> \(\frac{1}{2x+y+z}\le\frac{1}{16}.\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\) (*)
Tương tự, ta có: \(\frac{1}{x+2y+z}\le\frac{1}{16}.\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\) (**)
\(\frac{1}{x+y+2z}\le\frac{1}{16}.\left(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\) (***)
Từ (*)(**)(***) => Vế trái \(\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=\frac{1}{4}.\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{4}.4=1\)
=> đpcm
+) Áp dụng BĐT Cô - si cho 4 số dương x; x; y; z ta có:
x+x+y+z≥44√x.x.y.z
=> 2x + y + z ≥44√x.x.y.z (1)
Với 4 số dương 1x ;1x ;1y ;1z ta có: 1x +1x +1y +1z ≥4.4√1x .1x .1y .1z (2)
Từ (1)(2) => (2x+y+z)(1x +1x +1y +1z )≥4.4√x.x.y.z4.4√1x .1x .1y .1z =16
=> 12x+y+z ≤116 .(2x +1y +1z ) (*)
Tương tự, ta có: 1x+2y+z ≤116 .(1x +2y +1z ) (**)
1x+y+2z ≤116 .(1x +1y +2z ) (***)
Từ (*)(**)(***) => Vế trái ≤116 (4x +4y +4z )=14 .(1x +1y +1z )=14 .4=1
=> đpcm
Bổ đề:\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Ta có:\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)
Tương tự ta có:\(\dfrac{1}{2y+z+x}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)
\(\dfrac{1}{2z+x+y}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}\right)\)
Cộng vế với vế ta có:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{2y+z+x}+\dfrac{1}{2z+x+y}\le\dfrac{1}{16}\left[4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}.4.4=1\)
Dấu "=" xảy ra ⇔ \(x=y=z=\dfrac{3}{4}\)
\(ab+bc+ca\le a^2+b^2+c^2\le\frac{\left(a+b+c\right)^2}{3}\) ( bđt phụ + Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c\)
CM bđt phụ : \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow\)\(2x^2+2y^2+2z^2\ge2xy+2yz+2zx\)
\(\Leftrightarrow\)\(2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(\Leftrightarrow\)\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)
\(\Leftrightarrow\)\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( luôn đúng )
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)
Chúc bạn học tốt ~