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Lời giải:
Từ \(xy+yz+xz=xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\left(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\right)=(a,b,c)\Rightarrow a+b+c=1\)
Bài toán tương đương với việc chứng minh:
\(\frac{c^3}{(a+1)(b+1)}+\frac{a^3}{(b+1)(c+1)}+\frac{b^3}{(a+1)(c+1)}\geq \frac{1}{16}\)
Thật vậy, áp dụng BĐT AM-GM ta có:
\(\frac{c^3}{(a+1)(b+1)}+\frac{a+1}{64}+\frac{b+1}{64}\geq 3\sqrt[3]{\frac{c^3}{64^2}}=\frac{3c}{16}\)
Tương tự:
\(\frac{a^3}{(b+1)(c+1)}+\frac{b+1}{64}+\frac{c+1}{64}\geq \frac{3a}{16}\)
\(\frac{b^3}{(c+1)(a+1)}+\frac{c+1}{64}+\frac{a+1}{64}\geq \frac{3c}{16}\)
Cộng các BĐT thu được ở trên:
\(\Rightarrow \text{VT}+\frac{(a+b+c)+3}{32}\geq \frac{3}{16}(a+b+c)\)
\(\Leftrightarrow \text{VT}+\frac{1}{8}\geq \frac{3}{16}\Rightarrow \text{VT}\geq \frac{1}{16}\)
Ta có đpcm
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\Leftrightarrow x=y=z=3\)
Ta có \(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\frac{\left(yz+1\right)^2}{z^2}}{\frac{zx+1}{x}}+\frac{\frac{\left(zx+1\right)^2}{x^2}}{\frac{xy+1}{y}}+\frac{\frac{\left(xy+1\right)^2}{y^2}}{\frac{yz+1}{z}}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\ge\frac{\left(a_1+a_2+a_3\right)^2}{b_1+b_2+b_3}\)
Dấu "=" xảy ra khi \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{c_3}\)
\(P=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(P\ge a+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng BĐT: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
=> \(P\ge x+y+z+\frac{9}{x+y+z}=\left[x+y+z+\frac{9}{4\left(x+y+z\right)}\right]+\frac{27}{4\left(x+y+z\right)}\)
Ta có: \(x+y+z+\frac{9}{4\left(x+y+z\right)}\ge2\sqrt{\frac{9}{4}}=3;\frac{27}{4\left(x+y+z\right)}=\frac{27}{4\cdot\frac{3}{2}}=\frac{9}{2}\)
=> \(P\ge3+\frac{9}{2}=\frac{15}{2}\).
Dấu "=" xảy ra <=> x=y=z=\(\frac{1}{2}\)
Vậy MinP=\(\frac{15}{2}\)đạt được khi x=y=z=\(\frac{1}{2}\)
Ta có:
\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\left(\frac{yz+1}{z}\right)^2}{\left(\frac{zx+1}{x}\right)}+\frac{\left(\frac{zx+1}{x}\right)^2}{\left(\frac{xy+1}{y}\right)}+\frac{\left(\frac{xy+1}{y}\right)^2}{\left(\frac{yz+1}{z}\right)}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có:
\(\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)\(\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\ge\left(x+y+z\right)+\frac{9}{x+y+z}=\left(x+y+z\right)+\frac{9}{4\left(x+y+z\right)}\)
\(+\frac{27}{4\left(x+y+z\right)}\ge2\sqrt{\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{15}{2}\)(Áp dụng BĐT Cô - si cho 2 số không âm)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{2}\)
Ta đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c;\frac{1}{t}=d\) ( a, b, c, d >0 )
Khi đó ta cần chứng minh:
\(\frac{a^3}{\frac{1}{bc}+\frac{1}{cd}+\frac{1}{db}}+\frac{b^3}{\frac{1}{ac}+\frac{1}{cd}+\frac{1}{da}}+\frac{c^3}{\frac{1}{ab}+\frac{1}{bd}+\frac{1}{da}}+\frac{d^3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge\frac{1}{3}\left(a+b+c+d\right)\)
\(VT=\frac{a^3}{\frac{b+c+d}{bcd}}+\frac{b^3}{\frac{a+c+d}{acd}}+\frac{c^3}{\frac{a+b+d}{abd}}+\frac{d^3}{\frac{a+b+c}{abc}}\)
\(=\frac{a^3}{\frac{a\left(b+c+d\right)}{abcd}}+\frac{b^3}{\frac{b\left(a+c+d\right)}{abcd}}+\frac{c^3}{\frac{c\left(a+b+d\right)}{abcd}}+\frac{d^3}{\frac{d\left(a+b+c\right)}{abcd}}\)
\(=\frac{a^2}{b+c+d}+\frac{b^2}{a+c+d}+\frac{c^2}{a+b+d}+\frac{d^2}{a+b+c}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{3\left(a+b+c+d\right)}=\frac{a+b+c+d}{3}=VP\)
Vậy ta đã chứng minh được
\(\frac{a^3}{\frac{1}{bc}+\frac{1}{cd}+\frac{1}{db}}+\frac{b^3}{\frac{1}{ac}+\frac{1}{cd}+\frac{1}{da}}+\frac{c^3}{\frac{1}{ab}+\frac{1}{bd}+\frac{1}{da}}+\frac{d^3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge\frac{1}{3}\left(a+b+c+d\right)\)
Dấu "=" xảy ra <=> a = b = c = d
Vậy :
\(\frac{1}{x^3\left(yz+zt+ty\right)}+\frac{1}{y^3\left(xz+zt+tx\right)}+\frac{1}{z^3\left(xy+yt+tx\right)}+\frac{1}{t^3\left(xy+yz+zx\right)}\ge\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\)
Dấu "=" xảy ra <=> x = y = z = t = 1
\(xy+yz+zx=xyz\)
\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\) thì
\(\hept{\begin{cases}a+b+c=1\\P=\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\ge\frac{1}{16}\end{cases}}\)
Ta co:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{64}+\frac{1+c}{64}\ge\frac{3a}{16}\)
\(\Leftrightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}\ge\frac{3a}{16}-\frac{b}{64}-\frac{c}{64}-\frac{1}{32}\)
Từ đây ta co:
\(P\ge\left(a+b+c\right)\left(\frac{3}{16}-\frac{1}{64}-\frac{1}{64}\right)-\frac{3}{32}=\frac{1}{16}\)
\(\frac{3}{2}\ge x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(P\ge3\sqrt[3]{\frac{x\left(yz+1\right)^2.y\left(zx+1\right)^2.z\left(xy+1\right)^2}{z^2\left(zx+1\right)x^2\left(xy+1\right)y^2\left(yz+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}\)
Xét \(Q=\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}=\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{\sqrt{xy}.\sqrt{yz}.\sqrt{zx}}\)
Đặt \(\left(\sqrt{xy};\sqrt{yz};\sqrt{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c\le\frac{3}{2}\Rightarrow abc\le\frac{1}{8}\)
\(Q=\frac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}=\frac{1+a^2b^2c^2+a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2}{abc}\)
\(Q\ge\frac{1+a^2b^2c^2+3\sqrt[3]{a^2b^2c^2}+3\sqrt[3]{a^4b^4c^4}}{abc}=\frac{1}{abc}+abc+3\left(\frac{1}{\sqrt[3]{abc}}+\sqrt[3]{abc}\right)\)
\(Q\ge abc+\frac{1}{64abc}+3\left(\sqrt[3]{abc}+\frac{1}{4\sqrt[3]{abc}}\right)+\frac{63}{64abc}+\frac{9}{4\sqrt[3]{abc}}\)
\(Q\ge2\sqrt{\frac{abc}{64abc}}+6\sqrt{\frac{\sqrt[3]{abc}}{4\sqrt[3]{abc}}}+\frac{63}{64.\frac{1}{8}}+\frac{9}{4.\sqrt[3]{\frac{1}{8}}}=\frac{125}{8}\)
\(\Rightarrow P\ge3\sqrt[3]{Q}\ge3\sqrt[3]{\frac{125}{8}}=\frac{15}{2}\)
\(P_{min}=\frac{15}{2}\) khi \(a=b=c=\frac{1}{2}\) hay \(x=y=z=\frac{1}{2}\)
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)
\(\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)^3}\)
\(\Rightarrow3\left(xy+yz+zx\right)^3\le\left(\dfrac{9}{8}\right)^2\)
\(\Rightarrow\left(xy+yz+zx\right)^3\le\dfrac{27}{64}\)
\(\Rightarrow xy+yz+zx\le\dfrac{3}{4}\)