Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
ta có:\(P=\sum\dfrac{y^2z^2}{x\left(y^2+z^2\right)}=\sum\dfrac{\dfrac{1}{x}}{\dfrac{1}{y^2}+\dfrac{1}{z^2}}\)
đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\)thì giả thiết trở thành : \(a^2+b^2+c^2=1\).tìm Min \(P=\dfrac{a}{b^2+c^2}+\dfrac{b}{a^2+c^2}+\dfrac{c}{a^2+b^2}\)
ta có:\(\dfrac{a}{b^2+c^2}=\dfrac{a}{1-a^2}=\dfrac{a^2}{a\left(1-a^2\right)}\)
Áp dụng bất đẳng thức cauchy:
\(\left[a\left(1-a^2\right)\right]^2=\dfrac{1}{2}.2a^2\left(1-a^2\right)\left(1-a^2\right)\le\dfrac{1}{54}\left(2a^2+1-a^2+1-a^2\right)^3=\dfrac{4}{27}\)
\(\Rightarrow a\left(1-a^2\right)\le\dfrac{2}{3\sqrt{3}}\)\(\Rightarrow\dfrac{a^2}{a\left(1-a^2\right)}\ge\dfrac{3\sqrt{3}}{2}a^2\)
tương tự với các phân thức còn lại ta có:
\(P\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
hay \(x=y=z=\sqrt{3}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\\\dfrac{1}{z}=c\end{matrix}\right.\) Thì bài toán trở thành
Cho \(a^2+b^2+c^2=1\) tính GTNN của \(P=\dfrac{a}{b^2+c^2}+\dfrac{b}{c^2+a^2}+\dfrac{c}{a^2+b^2}\)
Ta có:
\(a^2+b^2+c^2=1\)
\(\Rightarrow a^2+b^2=1-c^2\)
\(\Rightarrow\dfrac{c}{a^2+b^2}=\dfrac{c^2}{c\left(1-c^2\right)}\)
Mà ta có: \(2c^2\left(1-c^2\right)\left(1-c^2\right)\le\dfrac{\left(2c^2+1-c^2+1-c^2\right)^3}{27}=\dfrac{8}{27}\)
\(\Rightarrow c\left(1-c^2\right)\le\dfrac{2}{3\sqrt{3}}\)
\(\Rightarrow\dfrac{c^2}{c\left(1-c^2\right)}\ge\dfrac{3\sqrt{3}c^2}{2}\)
\(\Rightarrow\dfrac{c}{a^2+b^2}\ge\dfrac{3\sqrt{3}c^2}{2}\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{b}{c^2+a^2}\ge\dfrac{3\sqrt{3}b^2}{2}\left(2\right)\\\dfrac{a}{b^2+c^2}\ge\dfrac{3\sqrt{3}a^2}{2}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) \(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
Ta có 1+x2 = xy + yz + xz +x2 = ( x+ z)(x+y)
TT : 1+y2 = (y+z)(y+x)
1+z2 = (z+x)(z+y)
⇒ P = 2
Vậy P =2
\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=8\)
=>\(8xyz=xyz+\sum x+\sum xy+1\)
=>\(\sum x^2+14xyz=\left(\sum x\right)^2+2\sum x+2\)
mặt khác
\(8=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\ge\dfrac{8}{\sqrt[3]{xyz}}\rightarrow xyz\ge1\)
đặt \(\sum x=a\left(a\ge3\right)\)
khi đó \(P=\dfrac{a^2+2a+2}{4a^2+15xyz}\le\dfrac{a^2+2a+2}{4a^2+15}\)
\(\dfrac{a^2+2a+2}{4a^2+15}=\dfrac{1}{3}-\dfrac{\left(a-3\right)^2}{12a^2+45}\le\dfrac{1}{3}\)
vậy max bằng 1/3 khi x=y=z=1
Ta có
\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)=1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xyz}\)
áp dụng bất đẳng thức CS ta có
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}=9\) ;
\(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\ge\dfrac{9}{xy+yz+xz}\)
ta có đánh giá : \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{1}{3}\)
\(xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{27}\Rightarrow\dfrac{1}{xyz}\ge27\)
\(\Rightarrow1+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xyz}\ge1+9+27+27=64\)
\(\Rightarrowđpcm\)
Hướng dẫn: đặt \(A=\dfrac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\dfrac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\dfrac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)
Khi đó \(F-A=x-y+y-z+z-x=0\Rightarrow F=A\)
\(\Rightarrow2F=F+A=\sum\dfrac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x^2+y^2\right)^2}{2\left(x^2+y^2\right)\left(x+y\right)}\ge\sum\dfrac{\left(x+y\right)^2\left(x^2+y^2\right)}{4\left(x^2+y^2\right)\left(x+y\right)}\)
\(\Rightarrow2F\ge\dfrac{x+y+z}{2}\Rightarrow F\ge\dfrac{x+y+z}{4}\)
Bạn tham khảo tại link sau:
Câu hỏi của Thiều Khánh Vi - Toán lớp 9 | Học trực tuyến
\(\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{1}{y+1}+\dfrac{1}{z+1}=1\\xyz\left(x+y+z\right)\left(x+1\right)\left(y+1\right)\left(z+1\right)=1296\end{matrix}\right.\)
Đặt \(\dfrac{1}{x+1}=a;\dfrac{1}{y+1}=b;\dfrac{1}{z+1}=c\left(a,b,c>0\right)\)
\(\Rightarrow a+b+c=1\)
\(\dfrac{1}{x+1}=a\)
\(\Rightarrow x+1=\dfrac{1}{a}\)
\(\Rightarrow x=\dfrac{1}{a}-1=\dfrac{1-a}{a}=\dfrac{b+c}{a}\)
Tương tự, ta có: \(y=\dfrac{a+c}{b};z=\dfrac{a+b}{c}\)
Đặt \(M=xyz\left(x+y+z\right)\left(x+1\right)\left(y+1\right)\left(z+1\right)\)
\(=\dfrac{\left(b+c\right)\left(a+c\right)\left(a+b\right)}{abc}\times\left(\dfrac{b+c}{a}+\dfrac{a+c}{b}+\dfrac{a+b}{c}\right)\times\dfrac{1}{abc}\)
\(=\dfrac{\left(b+c\right)\left(a+c\right)\left(a+b\right)}{a^2b^2c^2}\times\left(\dfrac{b}{a}+\dfrac{a}{b}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{c}\right)\)
\(\ge\dfrac{8abc}{a^2b^2c^2}\times\left(2+2+2\right)\) (bđt AM - GM)
\(\ge\dfrac{8}{\dfrac{\left(a+b+c\right)^3}{27}}\times6=1296\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\Rightarrow x=y=z=2\)
Chừ ms onl nên ko bt
Ta có: \(X=\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right)\)
\(=1+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xy}+\dfrac{1}{xyz}\right)\)
\(\ge1+\dfrac{9}{x+y+z}+\left(\dfrac{x+y+z}{xyz}+\dfrac{1}{xyz}\right)\)
\(=10+\dfrac{2}{xyz}\) ( Do \(x+y+z=1\) )
Áp dụng BĐT AM-GM ta có:
\(\left(\dfrac{x+y+z}{3}\right)^3\ge xyz\) \(\Leftrightarrow\dfrac{1}{xyz}\ge27\)
\(\Rightarrow X\ge10+27.2=64\)
\(\Rightarrow\) Dấu ''='' xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Mỹ Duyên : cho hỏi chút : sao biết \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)
mà có được \(1+\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\left(\dfrac{1}{yz}+\dfrac{1}{xz}+\dfrac{1}{xy}+\dfrac{1}{xyz}\right)\ge1+\dfrac{9}{x+y+x}+\left(\dfrac{x+y+z}{xyz}+\dfrac{1}{xyz}\right)\)