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Đặt cái ban đầu là P
Ta có: \(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta lại có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta có:
\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
thật vĩ đại Hung nguyen
1/
\(S=\dfrac{1}{x}+\dfrac{2^2}{y}\ge\dfrac{\left(1+2\right)^2}{x+y}=\dfrac{9}{1}=9\)
\(\Rightarrow S_{min}=9\) khi \(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{2}{y}\\x+y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)
2/
Áp dụng BĐT: \(2\left(x^2+y^2\right)\ge\left(x+y\right)^2\Rightarrow x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\)
\(\Rightarrow\dfrac{\left(x+y\right)^2}{2}-3\left(x+y\right)\le x^2+y^2-3\left(x+y\right)=-4\)
\(\Rightarrow\dfrac{\left(x+y\right)^2}{2}-3\left(x+y\right)+4\le0\Leftrightarrow\left(x+y\right)^2-6\left(x+y\right)+8\le0\)
Đặt \(x+y=a\Rightarrow a^2-6a+8\le0\Rightarrow2\le a\le4\)
\(\Rightarrow2\le x+y\le4\)
\(\Rightarrow S\in\left[2;4\right]\)
Lời giải:
Từ \(x+y-z=-1\Rightarrow z-x-y=1\)
Ta có các biến đổi sau:
\(x+yz=x(z-x-y)+yz=x(z-x)+y(z-x)=(x+y)(z-x)\)
\(=(x+y)(y+1)\)
\(y+zx=y(z-x-y)+zx=y(z-y)+x(z-y)=(y+x)(z-y)\)
\(=(y+x)(x+1)\)
\(z+xy=z(z-x-y)+xy=(z-x)(z-y)=(x+1)(y+1)\)
Khi đó:\(P=\frac{x^3y^3}{(x+y)^2(x+1)^3(y+1)^3}(*)\)
Áp dụng BĐT Cauchy:
\((x+y)^2\geq 4xy\)
\(x+1=\frac{x}{2}+\frac{x}{2}+1\geq 3\sqrt[3]{\frac{x^2}{4}}\Rightarrow (x+1)^3\geq \frac{27x^2}{4}\)
\(y+1\geq 3\sqrt[3]{\frac{y^2}{4}}\Rightarrow (y+1)^3\geq \frac{27y^2}{4}\) (tương tự ở trên)
\(\Rightarrow (x+y)^2(x+1)^3(y+1)^3\geq \frac{729}{4}x^3y^3(**)\)
Từ \((*); (**)\Rightarrow P\leq \frac{x^3y^3}{\frac{729}{4}x^3y^3}=\frac{4}{279}\Rightarrow P_{\max}=\frac{4}{729}\)
Đẳng thức xảy ra khi \(x=y=2; z=5\)
\(BĐT\Leftrightarrow\dfrac{x}{y^3}+\dfrac{y}{z^3}+\dfrac{z}{x^3}\ge x+y+z\)
Đặt \(\left\{{}\begin{matrix}a=\dfrac{1}{x}\\b=\dfrac{1}{y}\\c=\dfrac{1}{z}\end{matrix}\right.\) \(\Rightarrow abc\ge1\)
\(BĐT\Leftrightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(VT=\dfrac{a^4}{ab}+\dfrac{b^4}{bc}+\dfrac{c^4}{ac}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}=\dfrac{\left(ab+bc+ac\right)^2}{ab+bc+ac}=ab+bc+ac\)
Ta có \(abc\ge1\)
\(\Rightarrow\left\{{}\begin{matrix}bc\ge\dfrac{1}{a}\\ab\ge\dfrac{1}{c}\\ac\ge\dfrac{1}{b}\end{matrix}\right.\Rightarrow bc+ac+ab\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
\(\Leftrightarrow\dfrac{x\left(1-y^3\right)}{y^3}+\dfrac{y\left(1-z^3\right)}{z^3}+\dfrac{z\left(1-x^3\right)}{x^3}\ge0\)
\(\left(x^3+y^3\right)\left(x+y\right)=xy\left(1-x\right)\left(1-y\right)\Leftrightarrow\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)=\left(1-x\right)\left(1-y\right)\left(1\right)\)
Ta có : \(\left(\frac{x^2}{y}+\frac{y^2}{x}\right)\left(x+y\right)\ge4xy\)
và \(\left(1-x\right)\left(1-y\right)=1-\left(x+y\right)+xy\le1-2\sqrt{xy}+xy\)
\(\Rightarrow1-2\sqrt{xy}+xy\ge4xy\Leftrightarrow0\) <\(xy\le\frac{1}{9}\)
Dễ chứng minh : \(\frac{1}{1+x^2}+\frac{1}{1+y^2}\le\frac{1}{1+xy};\left(x,y\in\left(0;1\right)\right)\)
\(\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}\le\sqrt{2\left(\frac{1}{1+x^2}+\frac{1}{1+y^2}\right)}\le\sqrt{2\left(\frac{2}{1+xy}\right)}=\frac{2}{\sqrt{1+xy}}\)
\(3xy-\left(x^2+y^2\right)=xy-\left(x-y\right)^2\le xy\)
\(\Rightarrow P\le\frac{2}{\sqrt{1+xy}}+xy=\frac{2}{\sqrt{1+t}}+t\), \(\left(t=xy\right)\), (0<\(t\le\frac{1}{9}\)
Xét hàm số :
\(f\left(t\right)=\frac{2}{\sqrt{t+1}}+t\) , (0<\(t\le\frac{1}{9}\)
Đề sai. Bạn xem lại đề.
Lời giải:
Áp dụng BĐT AM-GM:
$(4xy)^2=(x+y)^2\geq 4xy$
$\Rightarrow 4xy\geq 1\Rightarrow xy\geq \frac{1}{4}$
Bây giờ, cho $x=2; y=\frac{2}{7}$ thỏa mãn điều kiện đề. Nhưng $xy=\frac{4}{7}>\frac{1}{3}$ nên tập giá trị $P=xy$ không thể là $[\frac{1}{4}; \frac{1}{3}]$ được.