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gọi P là cái 1/x+1/y+1/z nha
1) (1/x+1/y+1/z)^2 = 1/x^2 + 1/y^2 + 1/z^2 + 2/(xy) + 2/(yz) + 2/(zx)
---> 3 = P + 2(x+y+z)/(xyz) = P + 2 ---> P = 1
a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)
Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2
b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)
Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
Ta có \(\frac{\sqrt{x^2+2y^2}}{xy}=\sqrt{\frac{1}{y^2}+\frac{2}{x^2}}\)
Áp dụng BĐT Buniacoxki ta có
\(\sqrt{\left(\frac{1}{y^2}+\frac{2}{x^2}\right)\left(1+2\right)}\ge\sqrt{\left(\frac{1}{y}+\frac{2}{x}\right)^2}=\frac{1}{y}+\frac{2}{x}\)
=> \(\sqrt{3}A\ge3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3\)
=> \(A\ge\sqrt{3}\)
\(MinA=\sqrt{3}\)khi x=y=z=3
a/ \(\frac{2x+1}{\sqrt{x^2+2}}+\left(x+1\right)\left(\sqrt{1+\frac{2x+1}{x^2+2}}-1\right)+2x+1=0\)
\(\Leftrightarrow\frac{2x+1}{\sqrt{x^2+2}}+\frac{\left(x+1\right)\left(2x+1\right)}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+2x+1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\frac{1}{\sqrt{x^2+2}}+\frac{x+1}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+1\right)=0\)
\(\Rightarrow x=-\frac{1}{2}\)
b/ \(Q\ge\frac{\left(x+y+z\right)^2}{xyz\left(x+y+z\right)}+\frac{\left(x^3+y^3+z^3\right)^2}{xy+yz+zx}\ge\frac{x+y+z}{xyz}+\frac{\left(x^2+y^2+z^2\right)^3}{\left(x+y+z\right)^2}\)
\(Q\ge\frac{27\left(x+y+z\right)}{\left(x+y+z\right)^3}+\frac{\left(x+y+z\right)^6}{27\left(x+y+z\right)^2}=\frac{27}{\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}\)
\(Q\ge\frac{27}{64\left(x+y+z\right)^2}+\frac{27}{64\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}+\frac{837}{32\left(x+y+z\right)^2}\)
\(Q\ge3\sqrt[3]{\frac{27^2\left(x+y+z\right)^4}{64^2.27\left(x+y+z\right)^4}}+\frac{837}{32.\left(\frac{3}{2}\right)^2}=\frac{195}{16}\)
"=" \(\Leftrightarrow x=y=z=\frac{1}{2}\)
Nguyễn Trúc Giang, Duy Khang, Vũ Minh Tuấn, Võ Hồng Phúc, tth, No choice teen, Phạm Lan Hương,
Nguyễn Lê Phước Thịnh, @Nguyễn Việt Lâm, @Akai Haruma
giúp em vs ạ! Cần trước 5h chiều nay ạ
Thanks nhiều
a) \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2\left(y^2+\frac{1}{x^2}\right)\)
\(+\frac{1}{y^2}\left(y^2+\frac{1}{x^2}\right)=x^2y^2+2+\frac{1}{x^2y^2}\)
\(=2+\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\frac{255}{256x^2y^2}\)
Áp dụng BĐT Cauchy - Schwar cho 2 số không âm, ta được:
\(x^2y^2+\frac{1}{256x^2y^2}\ge2\sqrt{\frac{x^2y^2}{256x^2y^2}}=\frac{1}{8}\)
C/m được BĐT phụ: \(1=\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow16x^2y^2\le1\Leftrightarrow256x^2y^2\le16\Leftrightarrow\frac{255}{256x^2y^2}\ge\frac{255}{16}\)
\(\Rightarrow M\ge2+\frac{1}{8}+\frac{255}{16}=\frac{289}{16}\)
(Dấu "="\(\Leftrightarrow\hept{\begin{cases}x^2y^2=\frac{1}{256x^2y^2}\\x-y=0\end{cases}}\Leftrightarrow x=y=\frac{1}{2}\))
\(\frac{16}{3x+3y+2z}=\frac{16}{\left(x+y\right)+\left(y+z\right)+\left(z+x\right)+\left(x+y\right)1}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\)
Tương tự \(\frac{16}{3x+2y+3z}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+z}\)
\(\frac{16}{2x+3y+3z}\le\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{y+z}\)
Cộng vế theo vế ta có:
\(16\left(\frac{1}{3x+2y+3z}+\frac{1}{3x+3y+2z}+\frac{1}{2x+3y+3z}\right)\le4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=24\)
\(\Rightarrow\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\left(đpcm\right)\)
P/S:Có dùng S-vác ngược dấu ạ.ý tưởng tách mẫu là từ tth_new - Trang của tth_new - Học toán với OnlineMath nha !
Ta có:
\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\ge\frac{9}{2\left(x+y+z\right)}\)\(\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có: \(\frac{1}{2x+3y+3z}=\frac{\left(\frac{3}{4}+\frac{1}{4}\right)^2}{2\left(x+y+z\right)+y+z}\le\frac{9}{32\left(x+y+z\right)}+\frac{1}{16\left(y+z\right)}\)
Do đó:
\(\frac{1}{2x+3y+3z}+\frac{1}{2y+3x+3z}+\frac{1}{2z+3x+3y}\)
\(\le\frac{9}{32\left(x+y+z\right)}\cdot3+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(\le\frac{9}{32\cdot\frac{3}{4}}+\frac{1}{16}\cdot6=\frac{3}{2}\)(Đpcm)