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Thay giá trị x = y = z vô thì thấy VT > 2 nên nghi ngờ đề sai. B xem lại
Gọi \(T=...\)
\(T+3=\frac{\sqrt{x}}{\sqrt{y}+\sqrt{z}}+1+\frac{\sqrt{y}}{\sqrt{z}+\sqrt{x}}+1+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}+1\)
\(T+3=\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\left(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{z}+\sqrt{x}}\right)\)
\(\ge\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right).\frac{\left(1+1+1\right)^2}{2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}=\frac{9}{2}\)\(\Rightarrow\)\(T\ge\frac{9}{2}-3=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)
...
Đặt \(\hept{\begin{cases}\sqrt{x}=a\\\sqrt{y}=b\\\sqrt{z}=c\end{cases}\left(a,b,c>0\right)}\)
Đặt \(P=\frac{\sqrt{x}}{\sqrt{y}+\sqrt{z}}+\frac{\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\)
\(\Rightarrow P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(\Rightarrow P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)
\(P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)
\(P+3=\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{c+a}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}\)
\(2\left(P+3\right)=2.\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
\(2\left(P+3\right)=\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)
Áp dụng BĐT AM-GM ta có:
\(2\left(P+3\right)\ge3.\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.3.\sqrt[3]{\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}=9.\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.\frac{1}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\)
\(\left(\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ne0\right)\)
\(\Leftrightarrow P+3\ge4,5\)
\(\Leftrightarrow P\ge1,5\)
\(P=1,5\Leftrightarrow a=b=c\Leftrightarrow\sqrt{x}=\sqrt{y}=\sqrt{z}\Leftrightarrow x=y=z\)
Vậy \(P_{min}=1,5\Leftrightarrow x=y=z\)
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 :(
\(ĐK:x\ge1,y\ge2,z\ge3\)
\(PT\Leftrightarrow\sqrt{x-1}+\frac{1}{\sqrt{x-1}}+\sqrt{y-2}+\frac{1}{\sqrt{y-2}}+\sqrt{z-3}+\frac{1}{\sqrt{z-3}}=6\)
Theo bđt AM-GM thì \(VT\ge6\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x-1}=\frac{1}{\sqrt{x-1}}=1\\\sqrt{y-2}=\frac{1}{\sqrt{y-2}}=1\\\sqrt{z-3}=\frac{1}{\sqrt{z-3}}=1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=4\end{cases}}\)