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\(\dfrac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\le\dfrac{xyz}{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}=\dfrac{1}{8}\)
Dấu "=" xảy ra khi \(x=y=z\)
\(VT=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(=2+\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\)
Bài toán trở thành \(\frac{z}{x}+\frac{y}{x}+\frac{y}{z}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}\ge\frac{x+y+z}{3\sqrt{xyz}}\)
Áp dụng bất đẳng thức AM-GM:
\(\frac{z}{x}+\frac{z}{y}+\frac{z}{z}\ge3\sqrt[3]{\frac{z^3}{xyz}}=\frac{3z}{\sqrt[3]{xyz}}\)
Tương tự:
\(\frac{y}{x}+\frac{y}{z}+\frac{y}{y}\ge\frac{3y}{\sqrt[3]{xyz}}\)
\(\frac{x}{z}+\frac{x}{y}+\frac{x}{x}\ge\frac{3x}{\sqrt[3]{xyz}}\)
\(\Leftrightarrow VT+3\ge3+\frac{3}{\sqrt[3]{xyz}}\left(x+y+z\right)\)
\(\Leftrightarrow VT\ge\frac{3\left(x+y+z\right)}{\sqrt[3]{xyz}}\)\(\ge\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Is it true?
Đặt cái ban đầu là P
Ta có: \(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta lại có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta có:
\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
thật vĩ đại Hung nguyen
\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)
Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Rightarrow abc\ge1\)
\(BĐT\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)
Ta có \(abc\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}bc\ge\dfrac{1}{a}\\ab\ge\dfrac{1}{c}\\ac\ge\dfrac{1}{b}\end{matrix}\right.\Rightarrow bc+ac+ab\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)
Ta có nhận xét sau:
\(\dfrac{x+2}{x^3\left(y+z\right)}=\dfrac{1}{x^2\left(y+z\right)}+\dfrac{2}{x^3\left(y+z\right)}=\dfrac{yz}{zx+xy}+\dfrac{2\left(yz\right)^2}{zx+xy}\)
Tương tự với các phân thức còn lại
Ta đặt:
\(\left\{{}\begin{matrix}a=xy\\b=yz\\c=zx\end{matrix}\right.\)
\(\Rightarrow abc=1\) và \(a,b,c>0\)
Biểu thức P trở thành:
\(P=\Sigma_{cyc}\dfrac{a}{b+c}+2\Sigma_{cyc}\dfrac{a^2}{b+c}\)
Dễ thấy:
\(\Sigma_{cyc}\dfrac{a}{b+c}\ge\dfrac{3}{2}\) (Nesbit)
\(\Sigma_{cyc}\dfrac{a^2}{b+c}\ge\dfrac{a+b+c}{2}\ge\dfrac{3\sqrt[3]{abc}}{2}=\dfrac{3}{2}\)
Do đó:
\(P\ge\dfrac{3}{2}+2.\dfrac{3}{2}=\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Giúp mình với @Ace Legona, @Akai Haruma, @Neet, @Võ Đông Anh Tuấn, @Nguyễn Huy Tú