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Bài 1 :
Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)
Theo BĐT Cô - Si dưới dạng engel ta có :
\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)
Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)
Áp dụng bất đẳng thức Bunyakovsky:
\(NL^2=\left(\sqrt{4x+2\sqrt{x}+1}+\sqrt{4y+2\sqrt{y}+1}+\sqrt{4z+2\sqrt{z}+1}\right)^2\)
\(\le\left(1^2+1^2+1^2\right)\left(4x+2\sqrt{x}+1+4y+2\sqrt{y}+1+4z+2\sqrt{z}+1\right)\)
\(=3\left(4x+4y+4z\right)+3\left(2\sqrt{x}+2\sqrt{y}+2\sqrt{z}\right)+3\left(1+1+1\right)\)
\(=12\left(x+y+z\right)+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+9\)
\(=153+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
Mặt khác,theo Bunyakovsky: \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\le3\left(x+y+z\right)=36\)
\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}\le6\)
\(\Rightarrow153+6\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\le153+36=189\)
\(\Rightarrow NL\le\sqrt{189}\)
Dấu "=" xảy ra khi: \(x=y=z=4\)
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=3\)
Khi đó \(\frac{1}{4x^2+y^2+z^2}=\frac{1}{3x^2+x^2+y^2+z^2}\le\frac{1}{3x^2+3}\)
Viết lại BĐT cần chứng minh như sau:
\(\frac{1}{3x^2+3}+\frac{1}{3y^2+3}+\frac{1}{3z^2+3}\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{x^2+1}+\frac{1}{y^2+1}+\frac{1}{z^2+1}\le\frac{3}{2}\)
Ta có BĐT phụ \(\frac{1}{x^2+1}\le-\frac{1}{2}x+1\)
\(\Leftrightarrow-\frac{x\left(x-1\right)^2}{2\left(x^2+1\right)}\ge0\) *luôn đúng*
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{1}{y^2+1}\le-\frac{1}{2}y+1;\frac{1}{z^2+1}\le-\frac{1}{2}z+1\)
Cộng theo vế 3 BĐT trên ta có:
\(VT\le-\frac{1}{2}\left(x+y+z\right)+3=-\frac{3}{2}+3=\frac{3}{2}=VP\)
Xảy ra khi x=y=z=1
Cho mih hỏi bđt phụ đó là sao, có thể CM giùm mih đc hok
\(\sum\dfrac{x^4y}{x^2+1}=\sum\dfrac{x^3.\dfrac{1}{z}}{x^2+xyz}=\sum\dfrac{x^2}{z\left(x+yz\right)}=\sum\dfrac{x^2}{xz+1}\)
Áp dụng bất đẳng thức cauchy-schwarz:
\(Vt=\sum\dfrac{x^2}{xz+1}\ge\dfrac{\left(x+y+z\right)^2}{xy+yz+xz+3}\)
mà theo AM-GM: \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}=3\)
hay \(3\le xy+yz+xz\)
do đó \(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+xz\right)}=\dfrac{3}{2}\)
Dấu = xảy ra khi x=y=z=1
P/s: Câu này khoai
\(A=\sum\sqrt{4x+2\sqrt{x}+1}\)
\(Max_A=+\infty\)
\("="x=y=z=+\infty\)
\(4\left(xy+yz+xz\right)+x+y+z=9\)
Mặt khác ta có \(\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\Rightarrow xy+yz+xz\le\dfrac{1}{3}\left(x+y+z\right)^2\)
\(\Rightarrow\dfrac{4}{3}\left(x+y+z\right)^2+\left(x+y+z\right)\ge9\)
\(\Leftrightarrow\left[2\left(x+y+z\right)+\dfrac{3}{4}\right]^2\ge\dfrac{441}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2\left(x+y+z\right)+\dfrac{3}{4}\ge\dfrac{21}{4}\\2\left(x+y+z\right)+\dfrac{3}{4}\le\dfrac{-21}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+z\ge\dfrac{9}{4}\\x+y+z\le-3\end{matrix}\right.\) \(\Rightarrow\left(x+y+z\right)^2\ge\dfrac{81}{16}\)
Mà \(P=x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}\ge\dfrac{81}{16.3}=\dfrac{27}{16}\)
\(\Rightarrow P_{min}=\dfrac{27}{16}\) khi \(x=y=z=\dfrac{3}{4}\)
Sửa đề: \(T=\frac{x}{1+4y^2}+\frac{y}{1+4z^2}+\frac{z}{1+4x^2}\)
\(T=x-\frac{4xy^2}{1+4y^2}+y-\frac{4yz^2}{1+4z^2}+z-\frac{4zx^2}{1+4x^2}\)
\(T\ge x+y+z-\frac{4xy^2}{4y}-\frac{4yz^2}{4z}-\frac{4zx^2}{4x}\)
\(T\ge\frac{3}{2}-\left(xy+yz+zx\right)\ge\frac{3}{2}-\frac{1}{3}\left(x+y+z\right)^2=\frac{3}{4}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
P= \(2\sqrt{x}+1+2\sqrt{y}+1+2\sqrt{z}+1\)
\(P^2=4\left(x+y+z\right)+3\)
với x+y+z=12 ta có\(P^2=4\cdot12+3=51\)
P=\(\sqrt{51}\)
vậy GTLN của p là \(\sqrt{51}\)