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Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\geq \frac{16}{3x+3y+2z}\)
\(\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\geq \frac{16}{3x+2y+3z}\)
\(\frac{1}{z+y}+\frac{1}{z+y}+\frac{1}{x+z}+\frac{1}{x+y}\geq \frac{16}{2x+3y+3z}\)
Cộng theo vế:
\(\Rightarrow 4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\geq 16\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)
\(\Rightarrow \frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\leq \frac{4.6}{16}=\frac{3}{2}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
Ta có :
\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(2x+y+z\right)+\left(2y+x+z\right)}\)(1)
Áp dụng BĐT \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow\left(1\right)\le\dfrac{1}{4}\left(\dfrac{1}{x+y+x+z}+\dfrac{1}{y+x+y+z}\right)\le\dfrac{1}{4}\left(\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}\right)\right)\)
\(=\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)\)
tương tự với hai ông còn lại sau đó cộng lại ta được:
\(\Sigma\dfrac{1}{3x+3y+2z}\le\dfrac{24}{16}=\dfrac{3}{2}\)
Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)
\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)
CMTT và cộng theo vế:
\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\Sigma\dfrac{1}{2x+3y+3z}\le\Sigma\dfrac{1}{16}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(\Rightarrow P\le\dfrac{4}{16}\Sigma\left(\dfrac{1}{x+y}\right)=\dfrac{2017}{4}\)
Dấu " = " xảy ra khi \(x=y=z=\dfrac{3}{4034}\)
Trai Vô Đối câu này đề thi vô lớp 10 tỉnh Thanh Hóa ( tất cả thí sinh nek .... lúc nào rảnh mik đăng lên thử xem sao )
Câu a :
Ta có : \(\sqrt{5+3x}-\sqrt{5-3x}=a\)
\(\Leftrightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Leftrightarrow5+3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}+5-3x=a^2\)
\(\Leftrightarrow10-2\sqrt{25-9x^2}=a^2\)
\(\Leftrightarrow2\sqrt{25-9x^2}=10-a^2\)
\(\Leftrightarrow\sqrt{25-9x^2}=\dfrac{10-a^2}{2}\)
\(\Leftrightarrow25-9x^2=\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow9x^2=25-\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow3x=\sqrt{\dfrac{50-\left(a^2-10\right)^2}{2}}\)
\(\Leftrightarrow x=\dfrac{\sqrt{50-\left(a^2-10\right)^2}}{3\sqrt{2}}\)
\(P=\dfrac{3\sqrt{2}.\sqrt{10+2\sqrt{\dfrac{10-a^2}{2}}}}{\sqrt{50-\left(a^2-10\right)^2}}\)
Bạn tự rút gọn nữa nhé :))
Câu b : \(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-24}{z}\)
\(=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-12}{z}\)
\(=3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{z}\right)\le3-3\left[\dfrac{\left(1+1+2\right)^2}{12}\right]=-1\)
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:
\(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}=6\ge\frac{9}{2\left(x+y+z\right)}\)\(\Rightarrow x+y+z\ge\frac{3}{4}\)
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)
Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{\left(1+1+1+1\right)^2}{a+b+c+d}=\frac{16}{a+b+c+d}\)ta có :
\(\frac{16}{3x+3y+2z}\le\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\)
\(\frac{16}{3x+2y+3z}\le\frac{1}{x+z}+\frac{1}{x+z}+\frac{1}{x+y}+\frac{1}{y+z}\)
\(\frac{16}{2x+3y+3z}\le\frac{1}{y+z}+\frac{1}{y+z}+\frac{1}{x+y}+\frac{1}{x+z}\)
Cộng theo vế 3 đẳng thức trên ta được :
\(16.\left(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\right)\)
\(\le4.\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=4.6=24\)
\(\Rightarrow\)\(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\le\frac{3}{2}\)
Câu hỏi của NGUYỄN DOÃN ANH THÁI - Toán lớp 9 - Học toán với OnlineMath
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(S^2=(2x+3y)^2\leq (3x^2+2y^2)\left(\frac{4}{3}+\frac{9}{2}\right)\leq \frac{6}{35}(\frac{4}{3}+\frac{9}{2})=1\)
\(\Rightarrow S\leq 1\)
Vậy $S_{\max}=1$. Giá trị này đạt tại \(\left\{\begin{matrix} 3x^2+2y^2=\frac{6}{35}\\ \frac{3}{2}x=\frac{2}{3}y\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{4}{35}\\ y=\frac{9}{35}\end{matrix}\right.\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(S^2=(2x+3y)^2\leq (3x^2+2y^2)\left(\frac{4}{3}+\frac{9}{2}\right)\leq \frac{6}{35}(\frac{4}{3}+\frac{9}{2})=1\)
\(\Rightarrow S\leq 1\)
Vậy $S_{\max}=1$. Giá trị này đạt tại \(\left\{\begin{matrix} 3x^2+2y^2=\frac{6}{35}\\ \frac{3}{2}x=\frac{2}{3}y\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{4}{35}\\ y=\frac{9}{35}\end{matrix}\right.\)