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Áp dụng BĐT Mincopxki ta có:
\(VT=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\)
\(=\sqrt{\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}}+\sqrt{\left(y+\frac{z}{2}\right)^2+\frac{3z^2}{4}}+\sqrt{\left(x+\frac{z}{2}\right)^2+\frac{3z^2}{4}}\)
\(\ge\sqrt{\left(x+y+z+\frac{x+y+z}{2}\right)^2+\left(\frac{\sqrt{3}\left(x+y+z\right)}{2}\right)^2}\)
\(=\sqrt{\frac{9\left(x+y+z\right)^2}{4}+\frac{3\left(x+y+z\right)^2}{4}}\)
\(=\sqrt{3\left(x+y+z\right)^2}=\sqrt{3}\left(x+y+z\right)=VP\)
Ta có : Áp dụng BĐT Cauchy ba số ở mẫu ta được
\(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\dfrac{x}{\dfrac{y+z+1}{3}}+\dfrac{y}{\dfrac{x+z+1}{3}}+\dfrac{z}{\dfrac{x+y+1}{3}}=\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\)Thấy: \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}\left(?!\right)\)
Ta phải chứng minh:
\(\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{3}\)
\(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
Mà \(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}=\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\)
Theo C.B.S
\(\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Phải chứng minh
\(\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
\(\Leftrightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
Ta có : \(xy+yz+xz\le x^2+y^2+z^2=3\)
Theo C.B.S : \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le9\)
\(\Rightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
=> ĐPCM
ta sử dụng bđt :\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)(dk mọi abcd)
cái này cm dễ thôi. bunhia nha
ĐĂT :\(A=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\)
\(\Rightarrow A=\sqrt{\left(x+\frac{y}{2}\right)^2+\left(\frac{y\sqrt{3}}{2}\right)^2}+\sqrt{\left(y+\frac{z}{2}\right)^2+\left(\frac{z\sqrt{3}}{2}\right)^2}+\sqrt{\left(z+\frac{x}{2}\right)^2+\left(\frac{x\sqrt{3}}{2}\right)^2}\)
Áp dingj bđt trên ta được \(A\ge\sqrt{\left(x+\frac{y}{2}+y+\frac{z}{2}+z+\frac{x}{2}\right)^2+\left(\frac{x\sqrt{3}}{2}+\frac{y\sqrt{3}}{2}+\frac{z\sqrt{3}}{2}\right)^2}\)
\(\Rightarrow A\ge\sqrt{\frac{9}{4}\left(x+y+z\right)^2+\frac{3}{4}\left(x+y+z\right)^2}=\sqrt{3}\left(x+y+z\right)\)(dpcm)
Dấu = xảy ra khi và chỉ khi x=y=z
Áp dụng BĐT AM-GM ta có:
\(\sqrt[3]{yz}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{x}{\frac{y+z+1}{3}}=\frac{3x}{y+z+1}\)
Tương tự rồi cộng lại ta có:
\(VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)\)
\(=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)\)
\(\ge\frac{3\left(x^4+y^4+z^4\right)}{2\left(xy+yz+xz\right)+x+y+z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}\)
\(=x^2+y^2+z^2\ge xy+yz+xz=VP\)
Đẳng thức xảy ra khi \(x=y=z=1\)
Áp dụng BĐT AM-GM ta có:
\sqrt[3]{yz}\le\frac{y+z+1}{3}\Rightarrow\frac{x}{\sqrt[3]{yz}}\ge\frac{x}{\frac{y+z+1}{3}}=\frac{3x}{y+z+1}3yz≤3y+z+1⇒3yzx≥3y+z+1x=y+z+13x
Tương tự rồi cộng lại ta có:
VT\ge3\left(\frac{x}{y+z+1}+\frac{y}{x+z+1}+\frac{z}{x+y+1}\right)VT≥3(y+z+1x+x+z+1y+x+y+1z)
=3\left(\frac{x^2}{xy+yz+x}+\frac{y^2}{xy+yz+y}+\frac{z^2}{yz+xz+z}\right)=3(xy+yz+xx2+xy+yz+yy2+yz+xz+zz2)
\ge\frac{3\left(x^4+y^4+z^4\right)}{2\left(xy+yz+xz\right)+x+y+z}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}≥2(xy+yz+xz)+x+y+z3(x4+y4+z4)≥x2+y2+z2(x2+y2+z2)2
=x^2+y^2+z^2\ge xy+yz+xz=VP=x2+y2+z2≥xy+yz+xz=VP
Đẳng thức xảy ra khi x=y=z=1x=y=z=1
solution:
ta có: \(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\Leftrightarrow xyz\le1\)(theo BĐT cauchy cho 3 số )
\(\Rightarrow xy\le\dfrac{1}{z};yz\le\dfrac{1}{x};xz\le\dfrac{1}{y}\)
\(\Rightarrow\dfrac{x}{\sqrt[3]{yz}}\ge\dfrac{x}{\dfrac{1}{\sqrt[3]{x}}}=x\sqrt[3]{x}=\sqrt[3]{x^4}\)
tương tự ta có:\(\dfrac{y}{\sqrt[3]{xz}}\ge\sqrt[3]{y^4};\dfrac{z}{\sqrt[3]{xy}}\ge\sqrt[3]{z^4}\)
cả 2 vế các BĐT đều dương,cộng vế với vế:
\(S=\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\sqrt[3]{x^4}+\sqrt[3]{y^4}+\sqrt[3]{z^4}\)
Áp dụng BĐT bunyakovsky ta có:
\(\left(\sqrt[3]{x^4}+\sqrt[3]{y^4}+\sqrt[3]{z^4}\right)\left(x^2+y^2+z^2\right)\ge\left(\sqrt[3]{x^8}+\sqrt[3]{y^8}+\sqrt[3]{z^8}\right)^2=\left(x^2+y^2+z^2\right)^2\)
\(\Rightarrow S\ge x^2+y^2+z^2\)
đến đây ta lại có BĐT quen thuộc: \(x^2+y^2+z^2\ge xy+yz+xz\)
\(\Rightarrow S\ge xy+yz+xz\left(đpcm\right)\)
dấu = xảy ra khi và chỉ khi x=y=z mà x2+y2+z2=3 => x=y=z=1
*cách khác : Áp dụng BĐT cauchy - schwarz(bunyakovsky):
\(S=\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}=\dfrac{x^4}{x^3.\dfrac{1}{\sqrt[3]{x}}}+\dfrac{y^4}{y^3.\dfrac{1}{\sqrt[3]{y}}}+\dfrac{z^4}{z^3.\dfrac{1}{\sqrt[3]{z}}}\)
\(S\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2}=x^2+y^2+z^2\ge xy+yz+xz\)
\(\text{Cho 3 số dương x, y, z thỏa mãn }x+y+z=3\)
\(\text{Chứng minh rằng }T=\dfrac{x}{x+\sqrt{3x+yz}}+\dfrac{y}{y+\sqrt{3y+xz}}+\dfrac{z}{z+\sqrt{3z+xy}}\le1\)
➤➤➤Chứng minh:
➢ Áp dụng bất đẳng thức AM - GM
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}=\dfrac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}\left(\text{vì }x+y+z=3\right)=\dfrac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\le\dfrac{x}{x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}}=\dfrac{x}{x+\sqrt{xz}+\sqrt{xy}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
➢ Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
\(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
➢ Công vế theo vế 3 bất đẳng thức cùng chiều
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}+\dfrac{y}{y+\sqrt{3y+xz}}+\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
➢ \(\text{Đẳng thức xảy ra khi }x=y=z=1\)
➤ \(Max_T=1\Leftrightarrow x=y=z=1\)
(x+y+z)^2/3>=(x^2+y^2+z^2+2xy+2yz+2zx)/3>=3(xy+yz+zx)/3=xy+yz+zx(do x^2+y^2+z^2>=xy+yz+zx)(1)
(xy+yz)/2>=y√xz;(yz+zx)/2>=z√xy;(zx+xy)/2>=x√yz(BĐT Cô-si)
Cộng theo vế >>>xy+yz+zx>=y√xz+z√xy+x√yz(2)
Từ(1),(2) >>>đpcm
ta có: x^2+xy+y^2=3/4(x+y)^2+1/4(x-y)^2>=3/4(x+y)^2
suy ra căn (x^2+xy+y^2)>=căn 3/2(x+y)
cmtt ta có A>= căn3(x+y+z)=căn3 (A la ve trai)
dấu=xay ra khi va chi khi x=y=z=1/3