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Áp dụng BĐT Cauchy, ta có:
\(VT\ge3\sqrt[3]{\dfrac{x^2.y^2.z^2}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}=3\sqrt[3]{\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}}\)
Ta có: xyz=1 và x,y,z >0
\(\Rightarrow x\le1\Rightarrow x+1\le2\Rightarrow\dfrac{1}{x+1}\ge\dfrac{1}{2}\)
Tương tự \(\dfrac{1}{y+1}\ge\dfrac{1}{2}\)
\(\dfrac{1}{z+1}\ge\dfrac{1}{2}\)
\(\Rightarrow VT\ge3\sqrt[3]{\dfrac{1}{x+1}.\dfrac{1}{y+1}.\dfrac{1}{z+1}}=\dfrac{3}{2}\)
Đẳng thức xảy ra khi x=y=z=1
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y+1}{4}\ge2\sqrt{\dfrac{x^2}{4}}=x\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{z+1}+\dfrac{z+1}{4}\ge y\\\dfrac{z^2}{x+1}+\dfrac{x+1}{4}\ge z\end{matrix}\right.\)
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}+\dfrac{x+y+z}{4}+\dfrac{3}{4}\ge x+y+z\)
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{3\left(x+y+z\right)}{4}-\dfrac{3}{4}\) (1)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow\dfrac{3\left(x+y+z\right)}{4}\ge\dfrac{9}{4}\)
\(\Rightarrow\dfrac{3\left(x+y+z\right)}{4}-\dfrac{3}{4}\ge\dfrac{3}{2}=1,5\) (2)
Từ (1) và (2)
\(\Rightarrow\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge1,5\) (đpcm )
Dấu " = " xảy ra khi \(x=y=z=1\)
Sửa: =>\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{3\left(x+y+z\right)}{4}-\dfrac{3}{4}\left(1\right)\)
\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)
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 \(abc\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}bc\ge\dfrac{1}{a}\\ab\ge\dfrac{1}{c}\\ac\ge\dfrac{1}{b}\end{matrix}\right.\Rightarrow bc+ac+ab\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{x^4}{y+3z}+\dfrac{y+3z}{16}+\dfrac{1}{4}+\dfrac{1}{4}\ge4\sqrt[4]{\dfrac{x^4}{y+3z}\cdot\dfrac{y+3z}{16}\cdot\dfrac{1}{4}\cdot\dfrac{1}{4}}=x\)
\(\Rightarrow\dfrac{x^4}{y+3z}\ge x-\dfrac{y+3z}{16}-\dfrac{1}{2}\)
Tương tự cho 2 BĐT còn lại:
\(\dfrac{y^4}{z+3x}\ge y-\dfrac{z+3x}{16}-\dfrac{1}{2};\dfrac{z^4}{x+3y}\ge z-\dfrac{x+3y}{16}-\dfrac{1}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\ge\dfrac{3}{4}\left(x+y+z\right)-\dfrac{3}{2}\ge\dfrac{3}{4}\cdot3-\dfrac{3}{2}=\dfrac{3}{4}\)
Đẳng thức xảy ra khi \(x=y=z=1\)
Cách khác:
\(\dfrac{x^4}{y+3z}+\dfrac{y^4}{z+3x}+\dfrac{z^4}{x+3y}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{4\left(x+y+z\right)}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{4.\sqrt{3\left(x^2+y^2+z^2\right)}}=\dfrac{\sqrt{\left(x^2+y^2+z^2\right)^3}}{4\sqrt{3}}\)
\(\ge\dfrac{\sqrt{\left(xy+yz+zx\right)^3}}{4\sqrt{3}}\ge\dfrac{3\sqrt{3}}{4\sqrt{3}}=\dfrac{3}{4}\)
Dấu = xảy ra khi \(x=y=z=1\)
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 \(x=y=z=3\)
theo bđt cauchy schwarz ta có
\(\left\{{}\begin{matrix}\dfrac{2\sqrt{x}}{x^3+y^2}\le\dfrac{2\sqrt{x}}{2\sqrt{x^3y^2}}=\dfrac{1}{xy}\\\dfrac{2\sqrt{y}}{y^3+z^2}\le\dfrac{2\sqrt{y}}{2\sqrt{y^3z^2}}=\dfrac{1}{yz}\\\dfrac{2\sqrt{z}}{z^3+x^2}\le\dfrac{2\sqrt{z}}{2\sqrt{z^3y^2}}=\dfrac{1}{zy}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\le\dfrac{\dfrac{1}{x^2}+\dfrac{1}{y^2}}{2}+\dfrac{\dfrac{1}{y^2}+\dfrac{1}{z^2}}{2}+\dfrac{\dfrac{1}{z^2}+\dfrac{1}{x^2}}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\)\(\Rightarrow dpcm\)