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\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\)
CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\) ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)
\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
Mặt khác : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)
Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
Ta có x2-xy+y2=\(\left(\dfrac{x+y}{2}\right)^2+3\left(\dfrac{x-y}{2}\right)^2\)\(\ge\)\(\left(\dfrac{x+y}{2}\right)^2\)
=>\(\dfrac{\sqrt{x^2-xy+y^2}}{x+y+2z}\ge\dfrac{x+y}{2\left(x+y+2z\right)}\)(1) . Tương tự ...
Đặt \(\left\{{}\begin{matrix}y+z=a\\x+z=b\\x+y=c\end{matrix}\right.\)(a,b,c>0). Khi đó ta có :
S=\(\dfrac{1}{2}\left(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\right)\ge\dfrac{3}{4}\) (Netbit)
Áp dụng liên tiếp bất đẳng thức Mincopxki và bất đẳng thức Cauchy-Schwarz:
\(A=\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(A\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(A\ge\sqrt{4+\dfrac{81}{4}}=\sqrt{\dfrac{97}{4}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
\(B=\sqrt{x^2+\dfrac{1}{y^2}+\dfrac{1}{z^2}}+\sqrt{y^2+\dfrac{1}{z^2}+\dfrac{1}{x^2}}+\sqrt{z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B=\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+2\left(\dfrac{\left(1+1+1\right)^2}{x+y+z}\right)^2}\)
\(B\ge\sqrt{\left(x+y+z\right)^2+\dfrac{162}{\left(x+y+z\right)^2}}\)
\(B\ge\sqrt{4+\dfrac{162}{4}}=\sqrt{\dfrac{89}{2}}\)
Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\z\ge0\end{matrix}\right.\)
\(A=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+2}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{2\sqrt{z}}{\sqrt{xz}+2\sqrt{z}+2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz}.\sqrt{z}}{\sqrt{xz}+\sqrt{xyz}.\sqrt{z}+\sqrt{xyz}}\)
\(=\dfrac{1}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{\sqrt{yz}+\sqrt{y}+1}\)
\(=\dfrac{\sqrt{yz}+\sqrt{y}+1}{\sqrt{yz}+\sqrt{y}+1}=1\)
\(\Rightarrow\sqrt{A}=\sqrt{1}=1\)
Đặt \(\left(\sqrt{x};2\sqrt{y};3\sqrt{z}\right)=\left(a;b;c\right)\Rightarrow a;b;c\ge0\)
Ta có:
\(\dfrac{2}{a+b+c}-\dfrac{1}{ab+bc+ca}\le\dfrac{2}{a+b+c}-\dfrac{3}{\left(a+b+c\right)^2}=-3\left(\dfrac{1}{a+b+c}-\dfrac{1}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Đẳng thức xảy ra khi và chỉ khi: \(a=b=c=1\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{4}\\z=\dfrac{1}{9}\end{matrix}\right.\)
bài 3:
a, đặt x12=y9=z5=k
=>x=12k,y=9k,z=5k
ta có: ayz=20=> 12k.9k.5k=20
=> (12.9.5)k^3=20
=>540.k^3=20
=>k^3=20/540=1/27
=>k=1/3
=>x=12.1/3=4
y=9.1/3=3
z=5.1/3=5/3
vậy x=4,y=3,z=5/3
b,ta có: x5=y7=z3=x225=y249=z29
A/D tính chất dãy tỉ số bằng nhau ta có:
x5=y7=z3=x225=y249=z29=x2+y2−z225+49−9=58565=9
=>x=5.9=45
y=7.9=63
z=3*9=27
vậy x=45,y=63,z=27
\(P=\dfrac{\sqrt{2x^2+y^2}}{xy}+\dfrac{\sqrt{2y^2+z^2}}{yz}+\dfrac{\sqrt{2z^2+x^2}}{xz}\)
\(P=\sqrt{\dfrac{2x^2+y^2}{x^2y^2}}+\sqrt{\dfrac{2y^2+z^2}{y^2z^2}}+\sqrt{\dfrac{2z^2+x^2}{x^2z^2}}\)
\(P=\sqrt{\dfrac{2}{y^2}+\dfrac{1}{x^2}}+\sqrt{\dfrac{2}{z^2}+\dfrac{1}{y^2}}+\sqrt{\dfrac{2}{x^2}+\dfrac{1}{z^2}}\)
\(P\ge\sqrt{\left(\dfrac{\sqrt{2}}{x}+\dfrac{\sqrt{2}}{y}+\dfrac{\sqrt{2}}{z}\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}=3\)
hicc , lm vậy mà còn quát con IM