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bài này có lập được bảng biến thiên, nhưng chắc chưa học nên làm cách cơ bản
ta có \(\frac{x^2}{x^2+yz+x+1}\le\frac{x^2}{2x\sqrt{yz+1}+x}=\frac{x}{2\sqrt{yz+1}+1}\) dấu "=" xảy ra khi x2=yz+1
ta lại có \(2=x^2+y^2+z^2=\left(x+y+z\right)^3-2x\left(y+z\right)-2yz\ge\left(x+y+z\right)^3-\frac{\left(x+y+z\right)^2}{2}-2yz\)
\(\Rightarrow\left(x+y+z\right)^2\le4\left(1+yz\right)\Rightarrow x+y+z\le2\sqrt{1+yz}\)
\(\Rightarrow\frac{y+z}{x+y+z+1}=1-\frac{x+1}{x+y+z+1}\le1-\frac{x+1}{2\sqrt{yz+1}+1}\)
do đó \(P\le\frac{x}{2\sqrt{yz+1}+1}+1-\frac{x+1}{2\sqrt{yz+1}+1}-\frac{1+yz}{9}=1-\frac{1}{2\sqrt{yz+1}+1}-\frac{1+yz}{9}\)
\(\le1-\frac{1}{yz+1+1+1}-\frac{1+yz}{9}=\frac{11}{9}-\left(\frac{1}{yz+3}+\frac{yz+3}{9}\right)\le\frac{11}{9}-\frac{2}{3}=\frac{5}{9}\)
dấu "=" xảy ra khi \(\orbr{\begin{cases}x=1;y=1;z=0\\x=1;y=0;z=1\end{cases}}\)
\(3-P=1-\frac{x}{x+1}+1-\frac{y}{y+1}+1-\frac{z}{z+1}\)
\(=\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{9}{x+y+z+3}=\frac{9}{1+3}=\frac{9}{4}\)
\(\Rightarrow P\le\frac{3}{4}\)
Dấu "=" xảy ra tại \(x=y=z=\frac{1}{3}\)
AP DUNG BDT CAUCHY-SCHWAR : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)(DAU "=" XAY RA KHI \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\))
...Cauchy-Schwarz:
\(Q\ge\frac{\left(1+2+3\right)^2}{x+y+z}=\frac{36}{1}=36\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y+z=1\\\frac{1}{x}=\frac{2}{y}=\frac{3}{z}\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=y\\3y=2z\\z=3x\end{cases}}\)
Giải tiếp t cái dấu = :v
\(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow ab+bc+ca=1\)
\(\Rightarrow P\ge\frac{2a}{\sqrt{1+a^2}}+\frac{2b}{\sqrt{1+b^2}}+\frac{2c}{\sqrt{1+c^2}}\)
Áp dụng BĐT AM-GM: \(P=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\frac{b}{\sqrt{\left(b+c\right)\left(b+a\right)}}+\frac{c}{\sqrt{\left(c+a\right)\left(c+b\right)}}\)
\(\le a\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+b\left(\frac{1}{4\left(a+b\right)}+\frac{1}{a-b}\right)-c\left(\frac{1}{4\left(b+c\right)}+\frac{1}{a-c}\right)=\frac{9}{4}\)
Đẳng thức xảy ra khi \(\left(x;y;z\right)=\left(\frac{\sqrt{15}}{7};\sqrt{15};\sqrt{15}\right)\)
từ giả thiết ta suy ra \(\sqrt[3]{x^2y^2z^2}\ge3\)
lại có x2 + 2yz = x2 + yz + yz \(\ge\)3\(\sqrt[3]{x^2y^2z^2}\)\(\ge\)9
nên \(\frac{1}{x^2+2yz}\le\frac{1}{9}\)
tương tự với 2 số còn lại nên ta được P \(\le\frac{1}{3}\)
dấu "=" xảy ra khi x = y = z = \(\sqrt{3}\)
\(Q=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{4}{z+4}\right)\le3-\frac{16}{x+y+z+6}=\frac{1}{3}\)
dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\frac{1}{2};\frac{1}{2};-1\right)\)
Ta có: \(P=1-\frac{1}{x+1}+1-\frac{1}{y+1}+\frac{1}{z-1}=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng BĐT Bunhiacôpski ta có:
\(\left(1+x+1+y+1+z\right)\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\ge\left(1+1+1\right)^2=3^2=9\)
\(\Rightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{9}{3+x+y+z}=\frac{9}{4}\)
\(\Rightarrow A\le3-\frac{9}{4}=\frac{12}{4}-\frac{9}{4}=\frac{3}{4}\)
\(\Rightarrow Max_A=\frac{3}{4}\Leftrightarrow x=y=z=\frac{1}{3}\)
\(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
Thay \(x+y+z=1\)vào biểu thức
\(\Rightarrow P=\frac{x}{2x+y+z}+\frac{y}{x+2y+z}+\frac{z}{x+y+2z}\)
Áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\forall a,b>0\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{2x+y+z}=\frac{x}{x+y+x+z}\le\frac{x}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{y}{x+2y+z}=\frac{y}{x+y+y+z}\le\frac{y}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{z}{x+y+2z}=\frac{z}{x+z+y+z}\le\frac{z}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow VT\le\frac{x}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)+\frac{y}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)\(+\frac{z}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)
\(\Rightarrow VT\le\frac{x}{4\left(x+y\right)}+\frac{x}{4\left(x+z\right)}+\frac{y}{4\left(x+y\right)}+\frac{y}{4\left(y+z\right)}+\frac{z}{4\left(x+z\right)}\)\(+\frac{z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\frac{x}{4\left(x+y\right)}+\frac{y}{4\left(x+y\right)}+\frac{x}{4\left(x+z\right)}+\frac{z}{4\left(x+z\right)}+\frac{y}{4\left(y+z\right)}\)\(+\frac{z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\frac{x+y}{4\left(x+y\right)}+\frac{x+z}{4\left(x+z\right)}+\frac{y+z}{4\left(y+z\right)}\)
\(\Rightarrow VT\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)
\(\Rightarrow P\le\frac{3}{4}\)
Vậy \(P_{max}=\frac{3}{4}\)
Dấu " = " xảy ra khi \(x=y=z=\frac{1}{3}\)
Chúc bạn học tốt !!!