Tìm GTNN B=\(x^2+\frac{3}{2}y^2+\frac{5}{4}z^2+2xy-xz-yz+y-\frac{1}{2}\)
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Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
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ta có:
\(S\ge\frac{x^3}{x^2+y^2+\frac{x^2+y^2}{2}}+\frac{y^3}{y^2+z^2+\frac{y^2+z^2}{2}}+\frac{z^3}{z^2+x^2+\frac{z^2+x^2}{2}}\)
\(\Rightarrow S\ge\frac{2x^3}{3\left(x^2+y^2\right)}+\frac{2y^3}{3\left(y^2+z^2\right)}+\frac{2z^3}{3\left(z^2+x^2\right)}\Rightarrow\frac{3}{2}S\ge P=\frac{x^3}{x^2+y^2}+\frac{y^3}{y^2+z^2}+\frac{z^3}{z^2+x^2}\)
\(\Rightarrow P=x-\frac{xy^2}{x^2+y^2}+y-\frac{yz^2}{y^2+z^2}+z-\frac{zx^2}{z^2+x^2}\ge\left(x+y+z\right)-\left(\frac{xy^2}{2xy}+\frac{yz^2}{2yz}+\frac{zx^2}{2xz}\right)\)
\(=\left(x+y+z\right)-\frac{1}{2}\left(x+y+z\right)=\frac{9}{2}\)
\(\Rightarrow\frac{3}{2}S\ge\frac{9}{2}\Rightarrow S\ge3\)
Vậy Min S=3 khi x=y=z=3
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\(P=\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\)
Sử dụng bất đẳng thức AM-GM cho 3 số thực dương ta có :
\(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}.\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}.\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}}\)
\(=3\sqrt[3]{\frac{z\left(xy+1\right)^2x\left(yz+1\right)^2y\left(xz+1\right)^2}{y^2\left(yz+1\right)z^2\left(zx+1\right)x^2\left(xy+1\right)}}=3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}}\)
\(=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}=3\sqrt[3]{\frac{xy+1}{x}.\frac{yz+1}{y}.\frac{zx+1}{z}}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Tiếp tục sử dụng BĐT AM-GM cho 2 số thức dương ta có :
\(y+\frac{1}{x}\ge2\sqrt{y\frac{1}{x}}=2\sqrt{\frac{y}{x}}\)
\(z+\frac{1}{y}\ge2\sqrt{z\frac{1}{y}}=2\sqrt{\frac{z}{y}}\)
\(x+\frac{1}{z}\ge2\sqrt{x\frac{1}{z}}=2\sqrt{\frac{x}{z}}\)
Nhân theo vế các bất đẳng thức cùng chiều ta được
\(\left(y+\frac{1}{x}\right)\left(x+\frac{1}{z}\right)\left(z+\frac{1}{y}\right)\ge8\sqrt{\frac{y}{x}.\frac{x}{z}.\frac{z}{y}}=8\)
Khi đó \(3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(x+\frac{1}{z}\right)\left(z+\frac{1}{y}\right)}\ge3\sqrt[3]{8}=3.2=6\)
Dấu = xảy ra khi và chỉ khi \(x=y=z=\frac{1}{3}\)
Vậy MinP=1/3 đạt được khi x=y=z=1/3
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\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
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\(P=\frac{1}{1+xy}+\frac{1}{1+xz}+\frac{1}{1+yz}\ge\frac{9}{3+xy+xz+yz}\)
Lại có :\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow xy+yz+zx\le x^2+y^2+z^2\le3\)
\(\Rightarrow P\ge\frac{9}{3+3}=1.5\)
Dấu bằng xảy ra khi x=y=z=1
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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{9}{xy+yz+xz}(1)\)
\(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+xz}+\frac{1}{xy+yz+xz}\geq \frac{9}{x^2+y^2+z^2+2(xy+yz+xz)}=\frac{9}{(x+y+z)^2}=9(2)\)
Áp dụng hệ quả quen thuộc của BĐT AM-GM ta có:
\(3(xy+yz+xz)\leq (x+y+z)^2=1\Rightarrow xy+yz+xz\leq \frac{1}{3}\)
\(\Rightarrow \frac{7}{xy+yz+xz}\geq 21(3)\)
Từ \((1);(2);(3)\Rightarrow \frac{1}{x^2+y^2+z^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+xz}\geq 9+21=30\)Vậy $P_{\min}=30$. Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{9}{xy+yz+xz}(1)\)
\(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+xz}+\frac{1}{xy+yz+xz}\geq \frac{9}{x^2+y^2+z^2+2(xy+yz+xz)}=\frac{9}{(x+y+z)^2}=9(2)\)
Áp dụng hệ quả quen thuộc của BĐT AM-GM ta có:
\(3(xy+yz+xz)\leq (x+y+z)^2=1\Rightarrow xy+yz+xz\leq \frac{1}{3}\)
\(\Rightarrow \frac{7}{xy+yz+xz}\geq 21(3)\)
Từ \((1);(2);(3)\Rightarrow \frac{1}{x^2+y^2+z^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\geq \frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+xz}\geq 9+21=30\)Vậy $P_{\min}=30$. Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$
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Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\hept{\begin{cases}\sqrt{xy}\le\frac{x+y}{2}\\\sqrt{yz}\le\frac{y+z}{2}\\\sqrt{xz}\le\frac{x+z}{2}\end{cases}}\)
Cộng theo từng vế
\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{x+z}{2}\)
\(\Rightarrow1\le\frac{2\left(x+y+z\right)}{2}\)
\(\Rightarrow1\le x+y+z\)
\(\Rightarrow\frac{1}{2}\le\frac{x+y+z}{2}\left(1\right)\)
Ta có : \(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)
Áp dụng bất đẳng thức cộng mẫu số :
\(\Rightarrow A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)
\(\Rightarrow A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\ge\frac{x+y+z}{2}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\frac{1}{2}\le\frac{x+y+z}{2}\le\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)
\(\Rightarrow\frac{1}{2}\le\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{z+x}\)
Vậy GTNN của \(A=\frac{1}{2}\)
Dấu " = " xảy ra khi và chỉ khi \(x=y=z=\frac{1}{3}\)
Chúc bạn học tốt !!!
Ta có: \(1=\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\)
=> \(x+y+z\ge1\)
Có: \(A\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\ge\frac{1}{2}\)
Dấu "=" xảy ra <=> x = y = z =1/3
Vậy min A = 1/2 <=> x = y = z = 1/3