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\(VT=\frac{1}{x+y+z}+\frac{1}{3xyz}\ge2\sqrt{\frac{1}{3xyz\left(x+y+z\right)}}\ge2\sqrt{\frac{1}{\left(xy+yz+zx\right)^2}}=\frac{2}{xy+yz+zx}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
BĐT <=> \(\sqrt{\frac{x+yz}{xyz}}+\sqrt{\frac{y+xz}{xyz}}+\sqrt{\frac{z+xy}{xyz}}\ge1+\sqrt{\frac{1}{xy}}+\sqrt{\frac{1}{yz}}+\sqrt{\frac{1}{xz}}\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)
Khi đó \(a+b+c=1\)
BĐT <=>\(\sqrt{a+bc}+\sqrt{b+ac}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\)
Ta có \(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{\left(a+\sqrt{bc}\right)^2}=a+\sqrt{bc}\)
Khi đó \(VT\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=VP\)(ĐPCM)
Dấu bằng xảy ra khi x=y=z=3
BĐT cho tương đương với
\(\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)
Với \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z};a+b+c=1\)
Ta có:
\(\sqrt{a+bc}=\sqrt{a\left(a+b+c\right)+bc}\)
\(=\sqrt{a^2+a\left(b+c\right)+bc}\ge\sqrt{a^2+2a\sqrt{bc}+bc}=a+\sqrt{bc}\)
Tương tự
\(\sqrt{b+ca}\ge b+\sqrt{ca};\sqrt{c+ab}\ge c+\sqrt{ab}\)
Từ đó ta có đpcm
Dấu "=" xảy ra khi x=y=z=3
Đặt \(\left(x;y;z\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)
\(VT=\sum\frac{\sqrt{1+a^6+b^6}}{a^3b^3}\ge\sum\frac{\sqrt{3\sqrt[3]{a^6b^6}}}{a^3b^3}=\sqrt{3}\left(\frac{1}{a^2b^2}+\frac{1}{b^2c^2}+\frac{1}{c^2a^2}\right)\)
\(VT\ge\sqrt{3}.3\sqrt[3]{\frac{1}{a^2b^2.b^2c^2.c^2a^2}}=3\sqrt{3}\)
Dấu "=" xảy ra khi \(a=b=c=1\) hay \(x=y=z=1\)
Ta có: \(xy+yz+zx=xyz\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)ta có: \(a,b,c>0;a+b+c=1\)do đó 0<a,b,c<1
\(P=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+6\left(ab+bc+ca\right)\)
\(=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+2\left(a+b+c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\left(\frac{b^2}{a}-2b+a\right)+\left(\frac{c^2}{b}-2c+b\right)+\left(\frac{a^2}{c}-2a+c\right)-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\frac{\left(a-b\right)^2}{a}+\frac{\left(b-c\right)^2}{b}+\frac{\left(c-a\right)^2}{c}-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\frac{\left(1-a\right)\left(a-b\right)^2}{a}+\frac{\left(1-b\right)\left(b-c\right)^2}{b}+\frac{\left(1-c\right)\left(c-a\right)^2}{c}+3\ge3\)
Vậy GTNN của P=3
\(xyz=1\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{yz}\\xy=\dfrac{1}{z}\\\end{matrix}\right.\)
\(\dfrac{1}{1+x+xy}+\dfrac{1}{1+y+yz}+\dfrac{1}{1+z+zx}\\ =\dfrac{1}{1+x+xy}+\dfrac{xyz}{yz\left(\dfrac{1}{yz}+\dfrac{1}{z}+1\right)}+\dfrac{xyz}{z\left(\dfrac{1}{z}+1+x\right)}\\ =\dfrac{1}{1+x+xy}+\dfrac{x}{x+xy+1}+\dfrac{xy}{xy+1+x}\\ =\dfrac{1+x+xy}{1+x+xy}\\ =1\)
Đặt \(P=\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+xz}\)
ta dễ thấy rằng
\(P=\frac{z}{z+zx+xyz}+\frac{x}{x+xy+xyz}+\frac{y}{y+yz+xyz}=\frac{z}{1+z+xz}+\frac{x}{1+x+xy}+\frac{y}{1+y+yz}\)
Bằng cách nhân tương tự ta có : \(P=\frac{zy}{y+yz+xyz}+\frac{xz}{z+xz+xyz}+\frac{xy}{x+xy+xyz}=\frac{zy}{1+y+yz}+\frac{xz}{1+z+xz}+\frac{xy}{1+x+xy}\)
Cộng lại ta có : \(3P=\frac{1+x+xy}{1+x+xy}+\frac{1+y+yz}{1+y+yz}+\frac{1+z+xz}{1+z+xz}=3\text{ hay }P=1\)
vậy ta có điều phải chứng minh