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\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)
tương tự
\(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4}\);
\(\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\);
cộng vế với vế => đpcm
Dấu "=" xảy ra <=> x=y=z=1
Cho x, y, z >0 thoả mãn x+y+z=1. Cmr: \(\frac{x}{x+yz}+\frac{y}{y+xz}+\frac{z}{z+xy}\le\frac{9}{4}\)
\(VT=\sum\frac{x}{x\left(x+y+z\right)+yz}=\sum\frac{x}{\left(x+y\right)\left(x+z\right)}=\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(VT=\frac{2\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(VT=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)-xyz}\)
\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+\frac{1}{9}3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}-xyz}\)
\(VT\le\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\frac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)+xyz-xyz}=\frac{9}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
\(P=x+y+z+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge x+y+z+\frac{18}{x+y+z}\)
\(P\ge x+y+z+\frac{1}{x+y+z}+\frac{17}{x+y+z}\)
\(P\ge2\sqrt{\left(x+y+z\right)\frac{1}{\left(x+y+z\right)}}+\frac{17}{1}=19\)
\(P_{min}=19\) khi \(x=y=z=\frac{1}{3}\)
Đề bài sai bạn: ví dụ cho \(y=z=0\); \(x=4\) thì \(\frac{4}{6}\le\frac{1}{3}\) (vô lý)
2.
Áp dụng bất đẳng thức Bunhiacopxki :
\(\left(1+9^2\right)\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow82\cdot\left(x^2+\frac{1}{x^2}\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow\sqrt{82}\cdot\sqrt{x^2+\frac{1}{x^2}}\ge x+\frac{9}{x}\)
Tương tự ta cũng có :
\(\sqrt{82}\cdot\sqrt{y^2+\frac{1}{y^2}}\ge y+\frac{9}{y}\)
\(\sqrt{82}\cdot\sqrt{z^2+\frac{1}{z^2}}\ge z+\frac{9}{z}\)
Cộng theo vế của các bất đẳng thức ta được :
\(\sqrt{82}\cdot\left(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\right)\ge x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\)
\(\Leftrightarrow\sqrt{82}\cdot P\ge x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}\)(1)
Mặt khác áp dụng bất đẳng thức Cauchy ta có :
\(x+\frac{9}{x}+y+\frac{9}{y}+z+\frac{9}{z}=81x+\frac{9}{x}+81y+\frac{9}{y}+81z+\frac{9}{z}-80x-80y-80z\)
\(\ge2\sqrt{\frac{81x\cdot9}{x}}+2\sqrt{\frac{81y\cdot9}{y}}+2\sqrt{\frac{81z\cdot9}{z}}-80\left(x+y+z\right)\)
\(\ge2\sqrt{729}+2\sqrt{729}+2\sqrt{729}-80\cdot1\)
\(=82\) (2)
Từ (1) và (2) suy ra \(\sqrt{82}\cdot P\ge82\)
\(\Leftrightarrow P\ge\sqrt{82}\) ( đpcm )
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
1.
Áp dụng bất đẳng thức Cauchy :
\(\frac{a^2+1}{a}+\frac{b^2+1}{b}+\frac{c^2+1}{c}\)
\(=a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}\)
\(=9a+\frac{1}{a}+9b+\frac{1}{b}+9c+\frac{1}{c}-8a-8b-8c\)
\(\ge2\sqrt{\frac{9a}{a}}+2\sqrt{\frac{9b}{b}}+2\sqrt{\frac{9c}{c}}-8\left(a+b+c\right)\)
\(\ge3\cdot2\sqrt{9}-8=10\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
