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\(P=\dfrac{x-t}{y+t}+\dfrac{y-x}{z+x}+\dfrac{z-y}{t+y}+\dfrac{t-z}{x+z}\)
\(P=\dfrac{x+z-\left(y+t\right)}{y+t}+\dfrac{y+t-\left(z+x\right)}{z+x}=\dfrac{x+z}{y+t}+\dfrac{y+t}{z+x}-2\)
\(P\ge2\sqrt{\dfrac{\left(x+z\right)\left(y+t\right)}{\left(y+t\right)\left(x+z\right)}}-2=0\)
Dấu "=" xảy ra khi \(x+z=y+t\)
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\(xy+yz+zx=3xyz\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=3\)
Có \(\dfrac{1}{x+2y+3z}=\dfrac{1}{\left(x+y\right)+\left(y+z\right)+2z}\le\dfrac{1}{9}\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{2z}\right)\le\dfrac{1}{9}\left(\dfrac{1}{4x}+\dfrac{1}{4y}+\dfrac{1}{4y}+\dfrac{1}{4z}+\dfrac{1}{2z}\right)=\dfrac{1}{9}\left(\dfrac{1}{4x}+\dfrac{1}{2y}+\dfrac{3}{4z}\right)\)
Tương tự cx có: \(\dfrac{1}{y+2z+3x}\le\dfrac{1}{9}\left(\dfrac{1}{4y}+\dfrac{1}{2z}+\dfrac{3}{4x}\right)\);\(\dfrac{1}{z+2x+3y}\le\dfrac{1}{9}\left(\dfrac{1}{4z}+\dfrac{1}{2x}+\dfrac{3}{4y}\right)\)
Cộng vế với vế \(\Rightarrow\Sigma\dfrac{1}{x+2y+3z}\le\dfrac{1}{9}\left(\dfrac{1}{4}+\dfrac{1}{2}+\dfrac{3}{4}\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{2}\)
Dấu "=" xayra khi x=y=z=1
Vậy \(P_{max}=\dfrac{1}{2}\)
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\(\sum\dfrac{x^2}{y^2+yz+z^2}\ge\sum\dfrac{x^2}{y^2+\dfrac{y^2+z^2}{2}+z^2}=\dfrac{2}{3}\sum\dfrac{x^2}{y^2+z^2}\ge\dfrac{2}{3}.\dfrac{3}{2}=1\) (BĐT cuối là BĐT Netsbitt)
Câu b là bài IMO 2001 USA, em có thể tìm thấy rất nhiều lời giải
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\(1,\dfrac{1}{1+x}=1-\dfrac{1}{1+y}+1-\dfrac{1}{1+z}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Cmtt: \(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(1+x\right)\left(1+z\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(1+x\right)\left(1+y\right)}}\)
Nhân VTV
\(\Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge8\sqrt{\dfrac{x^2y^2z^2}{\left(1+x\right)^2\left(1+y\right)^2\left(1+z\right)^2}}\\ \Leftrightarrow\dfrac{1}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\ge\dfrac{8xyz}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\\ \Leftrightarrow8xyz\le1\Leftrightarrow xyz\le\dfrac{1}{8}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{2}\)
\(2,\\ a,2x^2+y^2-2xy=1\\ \Leftrightarrow\left(x-y\right)^2+x^2=1\\ \Leftrightarrow\left(x-y\right)^2=1-x^2\ge0\\ \Leftrightarrow x^2\le1\Leftrightarrow\sqrt{x^2}\le1\Leftrightarrow\left|x\right|\le1\)
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\(\dfrac{\sqrt{2}}{\sqrt{2x}.\sqrt{y+z}}\ge\dfrac{\sqrt{2}}{\dfrac{2x+y+z}{2}}=\dfrac{2\sqrt{2}}{2x+y+z}\)
\(\Rightarrow A\ge\sum\dfrac{2\sqrt{2}}{2x+y+z}=2\sqrt{2}\sum\dfrac{1}{2x+y+z}\ge2\sqrt{2}.\dfrac{9}{4\left(x+y+z\right)}=\dfrac{18\sqrt{2}}{4.18\sqrt{2}}=\dfrac{1}{4}\)
\(\Rightarrow A_{min}=\dfrac{1}{4}\) khi \(x=y=z=6\sqrt{2}\)
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\(P=\Sigma\dfrac{x}{x+yz}=\Sigma\dfrac{x}{x(x+y+z)+yz}=\Sigma\dfrac{x}{x^2+xy+xz+yz} \\=\Sigma\dfrac{x}{(x+y)(x+z)}=\dfrac{2(xy+yz+zx)}{(x+y)(y+z)(z+x)}\)
Bất đẳng thức phụ: \(\Pi(x+y)\ge\dfrac{8}{9}(\Sigma x)(\Sigma xy)\)
\(\Leftrightarrow \Sigma(x^2y+x^2z-2xyz)\ge0\) ( đúng do AM-GM )
Dấu ''='' xảy ra khi và chỉ khi: \(x=y=z\)
Áp dụng vào bài toán chính:
\(P\le\dfrac{2(xy+yz+zx)}{\dfrac{8}{9}(\Sigma x)(\Sigma xy)}=\dfrac{9}{4}\)
Dấu ''='' xảy ra khi và chỉ khi: \(x=y=z=\dfrac{1}{3}\)
Vậy \(\max P =\dfrac{9}{4} \) khi \(x=y=z=\dfrac{1}{3}\)
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Lời giải:
Áp dụng BĐT AM-GM:
$\frac{x^3}{y(x+z)}+\frac{y}{2}+\frac{x+z}{4}\geq \frac{3}{2}x$
Tương tự với các phân thức còn lại, cộng theo vế và rút gọn ta được:
$\Rightarrow P=\sum \frac{x^3}{y(x+z)}\geq \frac{x+y+z}{2}$
Tiếp tục áp dụng AM-GM:
$x+y\geq 2\sqrt{xy}$
$y+z\geq 2\sqrt{yz}$
$x+z\geq 2\sqrt{xz}$
$\Rightarrow x+y+z\geq \sqrt{xy}+\sqrt{yz}+\sqrt{xz}=1$
$\Rightarrow P\geq \frac{1}{2}$
Vậy $P_{\min}=\frac{1}{2}$ khi $x=y=z=\frac{1}{3}$
\(\dfrac{x^3}{y\left(x+z\right)}+\dfrac{y}{2}+\dfrac{x+z}{4}\ge\dfrac{3x}{2}\)
Tương tự và cộng lại:
\(P+x+y+z\ge\dfrac{3}{2}\left(x+y+z\right)\)
\(\Rightarrow P\ge\dfrac{1}{2}\left(x+y+z\right)\ge\dfrac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=\dfrac{1}{2}\)
Biểu thức này có min chứ không có max