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+\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
+\(3+2\left(xy+yz+zx\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\le9\)
\(\Rightarrow B=\frac{1}{1+\sqrt{3+2\left(xy+yz+zx\right)}}\ge\frac{1}{1+3}=\frac{1}{4}\)
+\(A=\frac{x^2}{y+2z}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}=\frac{x^4}{x^2y+2zx^2}+\frac{y^4}{y^2z+2xy^2}+\frac{z^4}{z^2x+2yz^2}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2z+z^2x+2\left(xy^2+yz^2+zx^2\right)}\)
Áp dụng bđt Bunhiacopxki
\(x^2y+y^2z+z^2x=x.xy+y.yz+z.zx\le\sqrt{x^2+y^2+z^2}.\sqrt{x^2y^2+y^2z^2+z^2x^2}\)
\(\le\sqrt{x^2+y^2+z^2}.\sqrt{\frac{\left(x^2+y^2+z^2\right)^2}{3}}=3\)
(áp dụng \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))
Tương tự: \(xy^2+yz^2+zx^2\le3\)
\(\Rightarrow B\ge\frac{3^2}{3+2.3}=1\)
\(VT=A+B\ge1+\frac{1}{4}=\frac{5}{4}=VP\)
\(P=\frac{x\left(x+y+z\right)+yz}{y+z}+\frac{y\left(x+y+z\right)+zx}{z+x}+\frac{z\left(x+y+z\right)+xy}{x+y}\)
\(P=\frac{\left(x+y\right)\left(x+z\right)}{y+z}+\frac{\left(x+y\right)\left(y+z\right)}{z+x}+\frac{\left(x+z\right)\left(y+z\right)}{x+y}\)
\(P\ge\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=2\left(x+y+z\right)=2\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
Bài này đơn giản thôi :))
\(\text{HPT}\Leftrightarrow \left\{\begin{matrix} \frac{x+y}{xy}=\frac{3}{2}\\ \frac{y+z}{yz}=\frac{2}{3}\\ \frac{x+z}{xz}=\frac{7}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=\frac{3}{2}\\ \frac{1}{y}+\frac{1}{z}=\frac{2}{3}\\ \frac{1}{x}+\frac{1}{z}=\frac{7}{6}\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{2}{x}=\frac{3}{2}+\frac{7}{6}-\frac{2}{3}\\ \frac{2}{y}=\frac{3}{2}+\frac{2}{3}-\frac{7}{6}\\ \frac{2}{z}=\frac{2}{3}+\frac{7}{6}-\frac{3}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\\ z=6\end{matrix}\right.\)
Vậy $(x,y,z)=(1,2,6)$ là nghiệm của hệ phương trình
Ta có \(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\frac{\left(yz+1\right)^2}{z^2}}{\frac{zx+1}{x}}+\frac{\frac{\left(zx+1\right)^2}{x^2}}{\frac{xy+1}{y}}+\frac{\frac{\left(xy+1\right)^2}{y^2}}{\frac{yz+1}{z}}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\ge\frac{\left(a_1+a_2+a_3\right)^2}{b_1+b_2+b_3}\)
Dấu "=" xảy ra khi \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{c_3}\)
\(P=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(P\ge a+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng BĐT: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
=> \(P\ge x+y+z+\frac{9}{x+y+z}=\left[x+y+z+\frac{9}{4\left(x+y+z\right)}\right]+\frac{27}{4\left(x+y+z\right)}\)
Ta có: \(x+y+z+\frac{9}{4\left(x+y+z\right)}\ge2\sqrt{\frac{9}{4}}=3;\frac{27}{4\left(x+y+z\right)}=\frac{27}{4\cdot\frac{3}{2}}=\frac{9}{2}\)
=> \(P\ge3+\frac{9}{2}=\frac{15}{2}\).
Dấu "=" xảy ra <=> x=y=z=\(\frac{1}{2}\)
Vậy MinP=\(\frac{15}{2}\)đạt được khi x=y=z=\(\frac{1}{2}\)
Ta có:
\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\left(\frac{yz+1}{z}\right)^2}{\left(\frac{zx+1}{x}\right)}+\frac{\left(\frac{zx+1}{x}\right)^2}{\left(\frac{xy+1}{y}\right)}+\frac{\left(\frac{xy+1}{y}\right)^2}{\left(\frac{yz+1}{z}\right)}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có:
\(\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)\(\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\ge\left(x+y+z\right)+\frac{9}{x+y+z}=\left(x+y+z\right)+\frac{9}{4\left(x+y+z\right)}\)
\(+\frac{27}{4\left(x+y+z\right)}\ge2\sqrt{\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{15}{2}\)(Áp dụng BĐT Cô - si cho 2 số không âm)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{2}\)
Áp dụng BĐT Cosi cho 2 sô dương ta có: \(x^2+yz\ge2x\sqrt{yz}\)
Tương tự: \(y^2+zx\ge2y\sqrt{zx};z^2+xy\ge2z\sqrt{xy}\)
Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được:
\(\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{zx}}+\frac{1}{2z\sqrt{xy}}\le\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\Leftrightarrow\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{xyz}\le\frac{x+y+z}{xyz}\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le x+y+z\)
\(\Leftrightarrow\frac{1}{2}\left(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\right)\ge0\)(luôn đúng)
Dấu "=" xảy ra khi \(x=y=z\)
Áp dụng BĐT Cosi cho 2 số dương ta có: \(x^2+yz\ge2\sqrt{x^2yz}=2x\sqrt{yz}\)
Tương tự: \(y^2+zx\ge2y\sqrt{zx},z^2+xy\ge2z\sqrt{xy}\)
Khi đó BĐT sẽ được chứng minh nếu ta chỉ ra được:
\(\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{zx}}+\frac{1}{2z\sqrt{xy}}\le\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(\Leftrightarrow\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{xyz}\le\frac{x+y+z}{xyz}\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\le x+y+z\)
\(\Leftrightarrow\frac{1}{2}\left(\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\right)\ge0\)(luôn đúng)
Vậy BĐT được chứng minh. Dấu "=" xảy ra khi \(x=y=z\)
Cách 2:
Ta chuẩn hóa xyz=1
BĐT viết lại là \(\frac{x}{x^3+1}+\frac{y}{y^3+1}+\frac{z}{z^3+1}\le\frac{1}{2}\left(x+y+z\right)\)
Ta sử dụng đánh giá
\(x-\frac{2x}{x^3+1}+\frac{3}{2}\ge\frac{9x^2}{2\left(x^2+x+1\right)}\)\(\Leftrightarrow\frac{\left(x-1\right)^2\left(2x^4+3x^2+7x+3\right)}{2\left(x^3+1\right)\left(x^2+x+1\right)}\ge0\)
Do vậy ta cần c/m \(\frac{x^2}{x^2+x+1}+\frac{y^2}{y^2+y+1}+\frac{z^2}{z^2+z+1}\ge1\)
ta có \(\left(x;y;z\right)\rightarrow\left(\frac{a^2}{bc};\frac{b^2}{ca};\frac{c^2}{ab}\right)\)
BĐT viết lại là \(\frac{a^4}{a^4+a^2bc+\left(bc\right)^2}+\frac{b^4}{b^4+b^2ca+\left(ca\right)^2}+\frac{c^4}{c^4+c^2ab+\left(ab\right)^2}\ge1\)
Theo bđt Cauchy-Schwarz ta có
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+abc\left(a+b+c\right)+\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2}\)
Theo bđt AM-GM ta có
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+2\left(ab\right)^2+2\left(bc\right)^2+2\left(ca\right)^2}=1\)
Dấu "=" xảy ra khi a=b=c=> x=y=z
đk: x;y;z dương nhé
áp dụng bđt cosi ta có:
\(x^2+yz>=2\sqrt{x^2yz}=2x\sqrt{yz};y^2+xz>=2\sqrt{y^2xz}=2y\sqrt{xz};z^2+xy=2\sqrt{z^2xy}=2z\sqrt{xy}\)
\(\Rightarrow\frac{1}{x^2+yz}< =\frac{1}{2x\sqrt{yz}};\frac{1}{y^2+xz}< =\frac{1}{2y\sqrt{xz}};\frac{1}{z^2+xy}< =\frac{1}{2z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(1\right)\)
áp dụng bđt cosi ta có:
\(\frac{1}{xy}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{xz}}=\frac{2}{x\sqrt{yz}};\frac{1}{xy}+\frac{1}{yz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{yz}}=\frac{2}{y\sqrt{xz}};\)
\(\frac{1}{yz}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{yz}\cdot\frac{1}{xz}}=\frac{2}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{xy}+\frac{1}{xz}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{yz}+\frac{1}{xz}=\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}>=\frac{2}{x\sqrt{yz}}+\frac{2}{y\sqrt{xz}}+\frac{2}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}>=\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)>=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(2\right)\)
từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}>=\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\left(đpcm\right)\)
dấu = xảy ra khi x=y=z
nhầm từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)