CV
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TN
28 tháng 3 2017
Hi ~! Mình xin slot trước :)
Giải
Dự đoán dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\) khi đó \(P=\frac{3\sqrt{3}}{4}\)
Ta sẽ chứng minh nó là GTNN của \(P\)
Ta có: \(x^2+xy+y^2=\frac{3\left(x+y\right)^2+\left(x-y\right)^2}{4}\ge\frac{3\left(x+y\right)^2}{4}\)
Do đó ta cần chứng minh
\(\frac{x+y}{4yz+1}+\frac{y+z}{4xz+1}+\frac{x+z}{4xy+1}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{x+y}{\left(y+z\right)^2+1}+\frac{y+z}{\left(x+z\right)^2+1}+\frac{x+z}{\left(x+y\right)^2+1}\ge\frac{3}{2}\)
Ta có: \(x+y+z=\frac{3}{2}\Rightarrow2x+2y+2z=3\)
\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(x+z\right)=2\)
Đặt \(\hept{\begin{cases}a=x+y\\b=y+z\\c=z+x\end{cases}}\) thì ta cần chứng minh
\(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\ge\frac{3}{2}\)\(\forall\hept{\begin{cases}a,b,c>0\\a+b+c=3\end{cases}}\)
Lại có: \(\frac{a}{b^2+1}=a-\frac{ab^2}{b^2+1}\ge a-\frac{ab}{2}\)
Tương tự ta cũng có: \(\frac{b}{c^2+1}\ge b-\frac{bc}{2};\frac{c}{a^2+1}\ge c-\frac{ac}{2}\)
Cộng theo vế các BĐT ta có: \(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\ge a-\frac{ab}{2}+b-\frac{bc}{2}+c-\frac{ac}{2}\)
\(=\left(a+b+c\right)-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)
BĐT đã được c/m vậy ta có \(P\ge\frac{3\sqrt{3}}{4}\Leftrightarrow x=y=z=\frac{1}{2}\)
8 tháng 9 2017
Áp dụng bđt Svacsơ ta có :
\(P=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{x^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)
ta lại có : \(\left(x^2+y^2+z^2\right)\left(y^2+z^2+x^2\right)\ge\left(xy+yz+zx\right)^2\)( bunhiacopxki )
\(\Rightarrow x^2+y^2+z^2\ge\left|xy+yz+xz\right|\ge xy+yz+xz\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3zx\)
\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)=3\)
\(\Rightarrow x+y+z\ge\sqrt{3}\)
\(\Rightarrow P\ge\frac{x+y+z}{2}\ge\frac{\sqrt{3}}{2}\) có GTNN là \(\frac{\sqrt{3}}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)
Vậy \(P_{min}=\frac{\sqrt{3}}{2}\) tại \(x=y=z=\frac{1}{\sqrt{3}}\)
26 tháng 4 2020
Ta có \(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\frac{\left(yz+1\right)^2}{z^2}}{\frac{zx+1}{x}}+\frac{\frac{\left(zx+1\right)^2}{x^2}}{\frac{xy+1}{y}}+\frac{\frac{\left(xy+1\right)^2}{y^2}}{\frac{yz+1}{z}}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\ge\frac{\left(a_1+a_2+a_3\right)^2}{b_1+b_2+b_3}\)
Dấu "=" xảy ra khi \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{c_3}\)
\(P=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(P\ge a+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng BĐT: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
=> \(P\ge x+y+z+\frac{9}{x+y+z}=\left[x+y+z+\frac{9}{4\left(x+y+z\right)}\right]+\frac{27}{4\left(x+y+z\right)}\)
Ta có: \(x+y+z+\frac{9}{4\left(x+y+z\right)}\ge2\sqrt{\frac{9}{4}}=3;\frac{27}{4\left(x+y+z\right)}=\frac{27}{4\cdot\frac{3}{2}}=\frac{9}{2}\)
=> \(P\ge3+\frac{9}{2}=\frac{15}{2}\).
Dấu "=" xảy ra <=> x=y=z=\(\frac{1}{2}\)
Vậy MinP=\(\frac{15}{2}\)đạt được khi x=y=z=\(\frac{1}{2}\)
KN
26 tháng 4 2020
Ta có:
\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\left(\frac{yz+1}{z}\right)^2}{\left(\frac{zx+1}{x}\right)}+\frac{\left(\frac{zx+1}{x}\right)^2}{\left(\frac{xy+1}{y}\right)}+\frac{\left(\frac{xy+1}{y}\right)^2}{\left(\frac{yz+1}{z}\right)}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có:
\(\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)\(\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\ge\left(x+y+z\right)+\frac{9}{x+y+z}=\left(x+y+z\right)+\frac{9}{4\left(x+y+z\right)}\)
\(+\frac{27}{4\left(x+y+z\right)}\ge2\sqrt{\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{15}{2}\)(Áp dụng BĐT Cô - si cho 2 số không âm)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{2}\)
\(GT\Leftrightarrow3x^2+y^2+z^2+\left(y+z\right)^2=2\)
Áp dụng BĐT bunyakovsky:\(y^2+z^2\ge\frac{1}{2}\left(y+z\right)^2\)
\(2\ge\frac{3}{2}\left(y+z\right)^2+3x^2\Leftrightarrow4\ge3\left(y+z\right)^2+6x^2=3\left[\left(y+z\right)^2+2x^2\right]\)
\(\left(2+1\right)\left[\left(y+z\right)^2+2x^2\right]\ge2\left(x+y+z\right)^2\)
\(\left(x+y+z\right)^2\le2\Leftrightarrow-\sqrt{2}\le x+y+z\le\sqrt{2}\)