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\(A=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)
Áp dụng BĐT AM-GM ta có:
\(A\le\frac{1+x-1}{x}+\frac{2+y-2}{2y}+\frac{3+z-3}{3z}=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=2\\\sqrt{z-3}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1=1\\y-2=2\\z-3=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
Vậy \(A_{max}=\frac{11}{6}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
Xin lỗi bạn. Bài đó mk lm sai rồi.
Sửa:
Áp dụng BĐT AM-GM ta có:
\(A=\frac{1.\sqrt{x-1}}{x}+\frac{\sqrt{2}.\sqrt{y-2}}{\sqrt{2}.y}+\frac{\sqrt{3}.\sqrt{z-3}}{\sqrt{3}.z}\le\frac{\frac{1+x-1}{2}}{x}+\frac{\frac{2+y-2}{2}}{\sqrt{2}.y}+\frac{\frac{3+z-3}{2}}{\sqrt{3}.z}=\frac{1}{2}+\frac{1}{2.\sqrt{2}}+\frac{1}{2.\sqrt{3}}\)\(=\frac{\sqrt{6}+\sqrt{3}+\sqrt{2}}{2.\sqrt{6}}\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-1}=1\\\sqrt{y-2}=\sqrt{2}\\\sqrt{z-3}=\sqrt{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x-1=1\\y-2=2\\z-3=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
Vậy \(A_{max}=\frac{\sqrt{6}+\sqrt{2}+\sqrt{3}}{2.\sqrt{6}}\)\(\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)
Bài 1: \(T=\sqrt{\frac{x^3}{x^3+8y^3}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)
\(=\frac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\frac{2y^2}{\sqrt{y\left[y^3+\left(x+y\right)^3\right]}}\)
\(=\frac{x^2}{\sqrt{\left(x^2+2xy\right)\left(x^2-2xy+4y^2\right)}}+\frac{2y^2}{\sqrt{\left(xy+2y^2\right)\left(x^2+xy+y^2\right)}}\)
\(\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2y^2+\left(x+y\right)^2}\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2x^2+4y^2}=1\)
\(\Rightarrow T\ge1\)
Bài 2:
[Toán 10] Bất đẳng thức | Page 5 | HOCMAI Forum - Cộng đồng học sinh Việt Nam
\(\frac{xy\sqrt{z-1}+xz\sqrt{y-2}+yz\sqrt{x-3}}{xyz}\\ =\frac{xy\sqrt{z-1}}{xyz}+\frac{xz\sqrt{y-2}}{xyz}+\frac{yz\sqrt{x-3}}{xyz}\\ =\frac{\sqrt{z-1}}{z}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{x-3}}{x}\\ =\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\)
Áp dụng BDT Cô-si với 2 số không âm:
\(\Rightarrow\frac{2\sqrt{z-1}}{2z}+\frac{2\sqrt{2}\sqrt{y-2}}{2\sqrt{2}y}+\frac{2\sqrt{3}\sqrt{x-3}}{2\sqrt{3}x}\\ \le\frac{1+\left(z-1\right)}{2z}+\frac{2+\left(y-2\right)}{2\sqrt{2}y}+\frac{3+\left(x-3\right)}{2\sqrt{3}x}\\ =\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}+\frac{\sqrt{2}}{4}+\frac{\sqrt{3}}{6}\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}z-1=1\\y-2=2\\x-3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}z=2\\y=4\\x=6\end{matrix}\right.\)
Vậy.......
ta có: \(\frac{\sqrt{2x^2+y^2}}{xy}=\sqrt{\frac{2}{y^2}+\frac{1}{x^2}}\)
Áp dụng BĐT bunyakovsky:\(\left(2+1\right)\left(\frac{2}{y^2}+\frac{1}{x^2}\right)\ge\left(\frac{2}{y}+\frac{1}{x}\right)^2\)
\(\Rightarrow\frac{2}{y^2}+\frac{1}{x^2}\ge\frac{1}{3}\left(\frac{2}{y}+\frac{1}{x}\right)^2\).....bla bla
áp dụng bdt cauchy -schửat dạng engel ta có
\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)\(\ge\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2}=\frac{1}{2}\)
(do \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\) bn tự cm nhé)
dau = xay ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)