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\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\)
\(\ge\frac{3x}{y+z+1}+\frac{3y}{x+z+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+xz+x}+\frac{3y^2}{xy+yz+y}+\frac{3z^2}{xz+yz+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x^2+y^2+z^2}\)
\(\ge\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Dấu "=" <=> x=y=z=1
1, A= y^3(1-y)^2 = 4/9 . y^3 . 9/4 (1-y)^2
= 4/9 .y.y.y . (3/2-3/2.y)^2
=4/9 .y.y.y (3/2-3/2.y)(3/2-3/2.y)
<= 4/9 (y+y+y+3/2-3/2.y+3/2-3/2.y)^5
=4/9 . 243/3125
=108/3125
Đến đó tự giải
Áp dụng BDT AM-GM ta có:\(VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)\)
\(\Rightarrow\frac{VT}{3}\ge\frac{x^2}{xy+xz+x}+\frac{y^2}{yz+yx+y}+\frac{z^2}{xz+zy+z}\)
\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+xy+z}\) (Cauchy-Schwarz)
Do \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)\(\Rightarrow\left(x+y+z\right)^2\le\left(x^2+y^2+z^2\right)^2\)
\(\Rightarrow x+y+z\le x^2+y^2+z^2\).Suy ra
\(2\left(xy+yz+xz\right)+x+y+z\le2\left(xy+yz+xz\right)+x^2+y^2+z^2=\left(x+y+z\right)^2\)
Suy ra \(\frac{VT}{3}\le\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\Rightarrow VT\ge3\) (điều phải chứng minh)
Dấu "=" xảy ra khi x=y=z=1
vì x2+y2+z2=1 mà x2+y2+z2>=xy+yz+xz suy ra 1>= xy+yz+xz
x2+y2+z2=1 suy ra (x-y)2=1-2xy-z2 ,(y-z)2=1-2yz-x2,(x-z)2=(x-z)2=1-2xz-y2
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]=\)
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)(do (x-y)2=1-2xy-z2(y-z)2=1-2yz-x2,(x-z)2=(x-z)2=1-2xz-y2)
theo bdt cosi ta có:
\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)
\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2z\sqrt{2xy}+2y\sqrt{2xz}+2x\sqrt{2yz}\right)]\)
\(\le\sqrt{3}+\frac{1}{2\sqrt{3}}[3-3\sqrt[3]{\left(2z\sqrt{2xy}.2y\sqrt{2xz}.2x\sqrt{2yz}\right)}\)
\(=\sqrt{3}+\frac{\sqrt{3}}{2}[1-2\sqrt{2}.\sqrt[3]{xyz^2}]\)\(=\sqrt{3}\left(1+\frac{1}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)=\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
suy ra
\(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\left(doxy+yz+xz\le1\right)\)
ta giả sử:
\(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\Leftrightarrow\sqrt{3}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\) mà \(\sqrt{3}>\frac{3}{2}\)
suy ra \(\frac{3}{2}\ge\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\)(luôn đúng) suy ra điều giả sử trên là đúng
hay \(3\sqrt[3]{xyz}\ge\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
mà \(\frac{x+y+z}{xy+yz+xz}\ge3.\sqrt[3]{xyz}\),\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)\(\le\sqrt{3}\left(\frac{3}{2}-\sqrt{2}.\sqrt[3]{xyz^2}\right)\)
suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[3-\left(2xy+z^2+2yz+x^2+2xz+y^2\right)]\)
suy ra \(\frac{x+y+z}{xy+yz+xz}\ge\)\(\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]\)(đpcm)
em mới có lớp 8, nếu em làm sai cho em xin lỗi nha anh
Ta có:
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Áp dụng BĐT Cosi ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\)
Cmtt:
\(\dfrac{y^3}{y\sqrt{1-y^2}}\ge2y^3\)
\(\dfrac{z^3}{z\sqrt{1-z^2}}\ge2z^3\)
\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}+\dfrac{y^3}{y\sqrt{1-y^2}}+\dfrac{z^3}{z\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\) (ĐPCM)
Áp dụng BĐT AM-GM cho 3 số không âm, ta có: \(0< \sqrt[3]{yz.1}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{3x}{y+z+1}\)
Làm tương tự với 2 hạng tử còn lại rồi cộng theo vế thì có:
\(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}\ge3\left(\frac{x}{y+z+1}+\frac{y}{z+x+1}+\frac{z}{x+y+1}\right)\)
\(=3\left(\frac{x^2}{xy+xz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{zx+yz+z}\right)\ge^{Schwartz}3.\frac{\left(x+y+z\right)^2}{x+y+z+2\left(xy+yz+zx\right)}\)
\(=3.\frac{x^2+y^2+z^2+2\left(xy+yz+zx\right)}{x+y+z+2\left(xy+yz+zx\right)}\ge9.\frac{xy+yz+zx}{\sqrt{3\left(x^2+y^2+z^2\right)}+2\left(x^2+y^2+z^2\right)}\)
\(=9.\frac{xy+yz+zx}{3+2.3}=xy+yz+zx\) => ĐPCM.
Dấu "=" xảy ra khi x=y=z=1.
Bài 32:
a) P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
= \(1+\sqrt{2}\)
b) Có: \(x^2-2y^2=xy\)
\(\Leftrightarrow x^2-y^2-y^2-xy=0\)
\(\Leftrightarrow\left(x-y\right)\left(x+y\right)-y\left(y+x\right)\)
\(\Leftrightarrow\left(x+y\right)\left(x-y-y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x-2y\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+y=0\\x-2y=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\x=2y\end{cases}}}\)
Thay x=-y ta có: Q=\(\frac{-y-y}{-y+y}\)=\(\frac{-2y}{0}\)(loại )
Thay x=2y ta có : Q=\(\frac{2y-y}{2y+y}=\frac{y}{3y}=\frac{1}{3}\)
đề sai bạn ơi, nhỡ may x=y=z=0 thì sao
ừ nhỉ phải là x3+y3+z3=1 bạn ạ