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\(\frac{x^3}{2x+3y+5z}+\frac{y^3}{2y+3z+5x}+\frac{z^3}{2z+3x+5y}\)
\(\Leftrightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3zy+5xy}+\frac{z^4}{2z^2+3xz+5yz}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{2x^2+2y^2+2z^2+8xy+8yz+8xz}\)
\(\Leftrightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
Xét \(\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
Áp dụng bất đẳng thức Cauchy cho 3 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}x^2+y^2\ge2\sqrt{x^2y^2}=2xy\\y^2+z^2\ge2\sqrt{y^2z^2}=2yz\\x^2+z^2\ge2\sqrt{x^2z^2}=2xz\end{matrix}\right.\)
Cộng từng vế:
\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)
\(\Rightarrow xy+yz+xz\le x^2+y^2+z^2\)
\(\Rightarrow8\left(xy+yz+xz\right)\le8\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(\Rightarrow\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\frac{x^2+y^2+z^2}{10}\)
Ta có: \(x^2+y^2+z^2\ge\frac{1}{3}\)
\(\Rightarrow\frac{x^2+y^2+z^2}{10}\ge\frac{1}{30}\)
\(\Rightarrow\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\ge\frac{1}{30}\)
Vì \(\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
\(\Rightarrow\frac{x^4}{2x^2+3xy+5xz}+\frac{y^4}{2y^2+3yz+5xy}+\frac{z^4}{2z^2+3xz+5yz}\ge\frac{1}{30}\)
\(\Leftrightarrow\frac{x^3}{2x+3y+5z}+\frac{y^3}{2y+3z+5x}+\frac{z^3}{2z+3x+5y}\ge\frac{1}{30}\) ( đpcm )
chịu chết
Bài 1:
Đặt \(\left(x+y;y+z;z+x\right)=\left(a;b;c\right)\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\)
\(P=\frac{1}{2a+b+c}+\frac{1}{a+b+2c}+\frac{1}{a+2b+c}\)
\(P=\frac{1}{a+a+b+c}+\frac{1}{a+b+c+c}+\frac{1}{a+b+b+c}\)
\(\Rightarrow P\le\frac{1}{16}\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{2}{c}+\frac{1}{a}+\frac{2}{b}+\frac{1}{c}\right)\)
\(\Rightarrow P\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{6}{4}=\frac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{2}\) hay \(x=y=z=\frac{1}{4}\)
Bài 2:
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-xy=5\\\left(x+y\right)\left(x^2+y^2-xy\right)=5x+15y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2-xy=5\\5\left(x+y\right)=5x+15y\end{matrix}\right.\)
\(\Rightarrow10y=0\Rightarrow y=0\)
Thay vào pt đầu: \(x^2=5\Rightarrow x=\pm\sqrt{5}\)
Vậy nghiệm của hệ là \(\left(x;y\right)=\left(\sqrt{5};0\right);\left(-\sqrt{5};0\right)\)
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 \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
\(\text{Áp dụng BĐT:}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge\frac{16}{a+b+c+d}\)
\(\frac{1}{3x+3y+2z}=\frac{1}{\left(x+y\right)+\left(x+y\right)+\left(x+z\right)+\left(y+z\right)}\le\frac{1}{16}\left(\frac{2}{x+y}+\frac{1}{x+z}+\frac{1}{y+z}\right)\)
\(\text{tương tự với các BĐT còn lại }\)
\(\Rightarrow\frac{1}{3x+3y+2z}+\frac{1}{3x+3z+2y}+\frac{1}{3y+3z+2x}\le\frac{1}{16}.\left(\frac{4}{x+z}+\frac{4}{x+y}+\frac{4}{y+z}\right)=\frac{1}{16}.24=\frac{3}{2}\left(đpcm\right)\)
Ta có \(\sqrt{3x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+z\right)\left(x+y\right)}\ge\sqrt{xy}+\sqrt{xz}\)(BĐT buniacoxki)
=>\(VT\le\frac{x}{x+\sqrt{xz}+\sqrt{xy}}+\frac{y}{y+\sqrt{yx}+\sqrt{yz}}+\frac{z}{z+\sqrt{zx}+\sqrt{yz}}\)
=> \(VT\le\frac{\sqrt[]{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)(ĐPCM)
Dấu bằng xảy ra khi a=b=c=1
Liên tục áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) và ta có:
\(\frac{1}{3x+3y+2x}=\frac{1}{2\left(x+y\right)+\left(x+y+2z\right)}\le\frac{1}{4}\left(\frac{1}{2\left(x+y\right)}+\frac{1}{\left(x+z\right)+\left(y+z\right)}\right)\le\frac{1}{8\left(x+y\right)}+\frac{1}{16}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)
Chứng minh tương tự tạ có:
\(\frac{1}{3x+2y+3z}\le\frac{1}{8\left(z+x\right)}+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)
\(\frac{1}{2x+3y+3z}\le\frac{1}{8\left(y+z\right)}+\frac{1}{16}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)\)
Suy ra \(VT\le\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)+\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{4}\)