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Câu hỏi của Trần Thành Phát Nguyễn - Toán lớp 9 - Học toán với OnlineMath
\(\sqrt{x^2+\frac{1}{x^2}}=\sqrt{\frac{9}{10}}\cdot\sqrt{\left(x^2+\frac{1}{x^2}\right)\left(\frac{1}{9}+1\right)}\ge\sqrt{\frac{9}{10}}\cdot\left(\frac{x}{3}+\frac{1}{x}\right)\)
Tương tự:\(\sqrt{y^2+\frac{1}{y^2}}\ge\sqrt{\frac{9}{10}}\left(\frac{y}{3}+\frac{1}{y}\right);\sqrt{z^2+\frac{1}{z^2}}\ge\sqrt{\frac{9}{10}}\left(\frac{z}{3}+\frac{1}{z}\right)\)
Cộng lại ta có:
\(LHS\ge\sqrt{\frac{9}{10}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{x+y+z}{3}\right)\ge\sqrt{\frac{9}{10}}\left(\frac{9}{x+y+z}+\frac{x+y+z}{3}\right)\)
\(=\sqrt{\frac{9}{10}}\cdot\left(\frac{x+y+z}{3}+\frac{1}{3\left(x+y+z\right)}+\frac{26}{3\left(x+y+z\right)}\right)\)
ai đó giúp em đoạn này với.Em cô si xong thấy không đúng ạ :(
\(\left(1.x+9.\frac{1}{y}\right)^2\le\left(1^2+9^2\right)\left(x^2+\frac{1}{y^2}\right)\Rightarrow\sqrt{x^2+\frac{1}{y^2}}\)
\(\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{y}\right)\)
\(TT:\sqrt{y^2+\frac{1}{z^2}}\ge\frac{1}{\sqrt{82}}\left(x+\frac{9}{z}\right);\sqrt{z^2+\frac{1}{x^2}}\ge\frac{1}{\sqrt{82}}\left(z+\frac{9}{x}\right)\)
\(S\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\right)\)
\(\ge\frac{1}{\sqrt{82}}\left(x+y+z+\frac{81}{x+y+z}\right)\)
\(=\frac{1}{\sqrt{82}}\left[\left(x+y+z+\frac{1}{x+y+z}\right)+\frac{80}{x+y+z}\right]\ge\sqrt{82}\)
Ta có: \(\frac{1}{2}.2x\left(1-x\right)\left(1-x\right)\le\frac{1}{2}\left[\frac{2x+1-x+1-x}{3}\right]^3=\frac{4}{27}\)
\(\Rightarrow\sqrt{x}\left(1-x\right)\le\frac{2\sqrt{3}}{9}\Rightarrow\frac{1}{\sqrt{x}\left(1-x\right)}\ge\frac{9}{2\sqrt{3}}\)
\(\Rightarrow\frac{\sqrt{x}}{1-x}\ge\frac{3\sqrt{3}}{2}x\). Thiết lập tương tự hai BĐT còn lại và cộng theo vế thu được đpcm.
Gọi \(\overrightarrow{1a}=\left(x;\frac{1}{x}\right);\overrightarrow{b}=\left(y;\frac{1}{y}\right);\overrightarrow{c}=\left(z;\frac{1}{z}\right)\)
Ta có:
\(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}=\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|+\left|\overrightarrow{c}\right|\)
\(\ge\left|\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\right|=\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)\(\ge\sqrt{1^2+\frac{9^2}{\left(x+y+z\right)^2}}\)
\(=\sqrt{1+81}=\sqrt{82}\)
Áp dụng BDT MInkopki
VT\(\ge\)\(\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}=\sqrt{82}\)
BDT minkopki
\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}+\sqrt{e^2+f^2}\ge\sqrt{\left(a+c+e\right)^2+\left(b+d+f\right)^2}\)
a/ \(VT\ge\frac{\left(\sqrt{b}+\sqrt{c}\right)^2}{2\sqrt{a}}+\frac{\left(\sqrt{c}+\sqrt{a}\right)^2}{2\sqrt{b}}+\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{2\sqrt{c}}\)
\(VT\ge\frac{\left(\sqrt{b}+\sqrt{c}+\sqrt{c}+\sqrt{a}+\sqrt{a}+\sqrt{b}\right)^2}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)
\(VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(VT\ge\sqrt{a}+\sqrt{b}+\sqrt{c}+3\sqrt[3]{\sqrt{abc}}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Dấu "=" xảy ra khi \(a=b=c=1\)
b/ \(VT=\sum\frac{x}{x+\sqrt{x\left(x+y+z\right)+yz}}=\sum\frac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}\)
\(VT\le\sum\frac{x}{x+\sqrt{xz}+\sqrt{xy}}=\sum\frac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Bài 1 :
Áp dụng BĐT Cô - si cho 2 số không âm ta có :
\(VT=\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\Sigma_{cyc}\sqrt{\frac{bc}{a}}\right)\)
\(\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge\left(\sqrt{\frac{ca}{b}}+\sqrt{\frac{ab}{c}}\right)+\left(\sqrt{\frac{ab}{c}}+\sqrt{\frac{bc}{a}}\right)+\left(\sqrt{\frac{bc}{a}}+\sqrt{\frac{ca}{b}}\right)\)
\(\Leftrightarrow\Sigma_{cyc}\frac{b+c}{\sqrt{a}}\ge2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\ge\sqrt{a}+\sqrt{b}+\sqrt{c}\)
\(+3\sqrt[6]{abc}=\sqrt{a}+\sqrt{b}+\sqrt{c}+3\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Áp dụng BĐT Mincopxki và AM - GM ta có :
\(P=\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\left(\frac{9}{x+y+z}\right)^2}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}+\frac{80}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{\sqrt[2]{\left(x+y+z\right)^2.\frac{1}{\left(x+y+z\right)^2}+80}}\)
\(\ge\sqrt{2+80}=\sqrt{82}\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
Chúc bạn học tốt !!!
mk hơi vội nên sai 1 số lỗi nhỏ bn tự sửa nhé
\(A=\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\)
Áp dụng Bđt MIncopxki ta có:
\(A\ge\sqrt{\left(x+y+\right)^2+\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{81}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{\left(x+y+z\right)^2+\frac{1}{\left(x+y+z\right)^2}+\frac{80}{\left(x+y+z\right)^2}}\)
\(\ge\sqrt{2+80}=\sqrt{82}\)
Dấu = khi \(x=y=z=\frac{1}{3}\)