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\(A=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\ge\frac{9}{3+xy+yz+zx}\)
\(\ge\frac{9}{3+x^2+y^2+z^2}\ge\frac{9}{3+3}=\frac{3}{2}\)
\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
b) Ta có \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{y+z+z+x+x+y}\)(BĐT Schwarz)
\(=\frac{x+y+z}{2}=\frac{2}{2}=1\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^2}{y+z}=\frac{y^2}{z+x}=\frac{z^2}{x+y}\\x+y+z=2\end{cases}}\Leftrightarrow x=y=z=\frac{2}{3}\)
a) Có \(P=1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)
\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(BĐT Bunyakovsky)
\(=\sqrt{3.\left[2\left(x+y+z\right)+xy+yz+zx\right]}\)
\(\le\sqrt{3\left[4+\frac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3\left(4+\frac{4}{3}\right)}=4\)
Dấu "=" xảy ra <=> x = y = z = 2/3
Câu 1:
\(y^2+yz+z^2=1-\frac{3x^2}{2}\)
\(\Leftrightarrow2y^2+2yz+2z^2=2-3x^2\)
\(\Leftrightarrow\left(y+z\right)^2+y^2+z^2+3x^2=2\)
\(\Leftrightarrow\left(y+z\right)^2+x^2+2x\left(y+z\right)+y^2+z^2+2x^2-2x\left(y+z\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2=2-\left(x-y\right)^2-\left(x-z\right)^2\)
\(\Leftrightarrow A^2=2-\left[\left(x-y\right)^2+\left(x-z\right)^2\right]\le2\forall x;y;z\)
\(\Leftrightarrow-\sqrt{2}\le A\le\sqrt{2}\)
Vậy \(A_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x=y=z\\x+y+z=-\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow x=y=z=\frac{-\sqrt{2}}{3}\)
\(A_{max}=\sqrt{2}\Leftrightarrow a=b=c=\frac{\sqrt{2}}{3}\)
Câu 2:
Áp dụng BĐT Cauchy-Schwarz:
\(P=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\ge\frac{9}{3+xy+yz+zx}\ge\frac{9}{3+x^2+y^2+z^2}=\frac{9}{6}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Câu 3:
\(P=\frac{ab\sqrt{c-2}+bc\sqrt{a-3}+ca\sqrt{b-4}}{abc}\) ( \(a\ge3;b\ge4;c\ge2\) )
\(P=\frac{\sqrt{c-2}}{c}+\frac{\sqrt{a-3}}{a}+\frac{\sqrt{b-4}}{b}\)
Áp dụng BĐT Cauchy:
\(P=\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}\cdot\sqrt{c-2}}{c}+\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}\cdot\sqrt{a-3}}{a}+\frac{1}{2}\cdot\frac{2\cdot\sqrt{b-4}}{b}\)
\(\le\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\cdot\frac{2+c-2}{c}+\frac{1}{\sqrt{3}}\cdot\frac{1}{2}\cdot\frac{3+a-3}{a}+\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{4+b-4}{b}=\frac{1}{2}\cdot\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{2}\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=8\\c=4\end{matrix}\right.\)
Câu 4:
Đặt \(\sqrt{x}=a;\sqrt{y}=b\left(a;b\ge0\right)\)
\(M=a^2-2ab+3b^2-2a+1\)
\(M=a^2-a\left(2b+2\right)+3b^2+1\)
\(\Delta=\left(2b+2\right)^2-4\left(3b^2+1\right)\)
\(=-8b^2+8b\)
\(=-8b\left(b+1\right)\ge0\)
Vì \(b\ge0\) nên \(-8b\left(b+1\right)\le0\)
Dấu "=" xảy ra \(\Leftrightarrow b=0\)
Khi đó \(M=a^2-2a+1=\left(a-1\right)^2\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow a=1\)
Vậy \(M_{min}=1\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=0\end{matrix}\right.\)
Áp dụng bđt AM-GM ta có
\(P\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2.\left(yz+1\right)^2.\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=A\)
Ta có \(A=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng bđt AM-GM ta có
\(A\ge3\sqrt[3]{8\sqrt{\frac{xyz}{xyz}}}=3.2=6\)
\(\Rightarrow P\ge6\)
Dấu "=" xảy ra khi x=y=z=\(\frac{1}{2}\)
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\(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}+\frac{1}{4x}\right)\left(z+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4y}\right)\left(x+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4z}\right)}\)
\(\ge3\sqrt[3]{5\sqrt[5]{\frac{y}{256x^4}}\cdot5\sqrt[5]{\frac{z}{256y^4}}\cdot5\sqrt[5]{\frac{x}{256z^4}}}\)
\(=3\sqrt[3]{125\sqrt[5]{\frac{xyz}{256^3\left(xyz\right)^4}}}\)
\(=15\sqrt[3]{\sqrt[5]{\frac{1}{256^3\left(xyz\right)^3}}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\left(\frac{x+y+z}{3}\right)^9}}\)
\(\ge15\sqrt[15]{\frac{1}{256^3\cdot\frac{1}{2^9}}}=\frac{15}{2}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
Bài này thì AM-GM thôi
\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}\)
Sử dụng BĐT AM-GM cho 3 số không âm ta có :
\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)^2}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(zx+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)
\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{zx+1}{z}\right)}\)
\(=3\sqrt[3]{\left(\frac{xy}{x}+\frac{1}{x}\right)\left(\frac{yz}{y}+\frac{1}{y}\right)\left(\frac{zx}{z}+\frac{1}{z}\right)}=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Tiếp tục sử dụng AM-GM cho 2 số không âm ta được :
\(3\sqrt[3]{\left(2\sqrt[2]{y\frac{1}{x}}\right)\left(2\sqrt[2]{z\frac{1}{y}}\right)\left(2\sqrt[2]{x\frac{1}{z}}\right)}\ge3\sqrt[3]{\left(2\sqrt{\frac{y}{x}}\right)\left(2\sqrt{\frac{z}{y}}\right)\left(2\sqrt{\frac{x}{z}}\right)}\)
\(=3\sqrt[3]{8\left(\sqrt{\frac{y}{x}}.\sqrt{\frac{z}{y}}.\sqrt{\frac{x}{z}}\right)}=3\sqrt[3]{8.\sqrt{\frac{xyz}{xyz}}}=3\sqrt[3]{8}=3.2=6\)
Đẳng thức xảy ra khi và chỉ khi \(x=y=z=\frac{1}{2}\)
Vậy \(Min_P=6\)đạt được khi \(x=y=z=\frac{1}{2}\)
Theo hệ quả của bất đẳng thức Cauchy - Schwarz
\(\Rightarrow x^2+y^2+z^2\ge xy+yz+xz\)
Mà \(x^2+y^2+z^2\le3\)
\(\Rightarrow xy+yz+xz\le3\)
Ta có \(P=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+xz}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow P\ge\frac{\left(1+1+1\right)^2}{xy+1+yz+1+xz+1}=\frac{9}{xy+yz+xz+3}\left(1\right)\)
Ta có : \(xy+yz+xz\le3\)
\(\Rightarrow xy+yz+xz+3\le6\)
\(\Rightarrow\frac{9}{xy+yz+xz+3}\ge\frac{9}{6}=\frac{3}{2}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow P\ge\frac{3}{2}\)
Vậy \(P_{min}=\frac{3}{2}\)
Dấu " = " xảy ra khi \(x=y=z=1\)
Chúc bạn học tốt !!!