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Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)
Từ giả thiết suy ra : \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\)
Nên ta có : \(\frac{\sqrt{1+x^2}}{x}=\sqrt{\frac{1}{x^2}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}\right)\left(\frac{1}{x}+\frac{1}{z}\right)}\le\frac{1}{2}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " \(\Leftrightarrow y=z\)
Vậy \(\frac{1+\sqrt{1+x^2}}{x}\le\frac{1}{2}\left(\frac{4}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có :
\(\frac{1+\sqrt{1+y^2}}{y}\le\frac{1}{2}\left(\frac{1}{x}+\frac{4}{y}+\frac{1}{z}\right);\frac{1+\sqrt{1+z^2}}{z}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{4}{z}\right)\)
Vậy ta có :
\(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Dấu " = " \(\Leftrightarrow x=y=z\)
Ta có :
\(\left(x+y+z\right)^2-3\left(xy+yz+xx\right)=...=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\ge0\)
Nên \(\left(x+y+x\right)^2\ge3\left(xy+yz+xx\right)\)
\(\Rightarrow\left(xyz\right)^2\ge3\left(xy+yz+xz\right)\Rightarrow3\frac{xy+yz+xz}{xyz}\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le xyz\)
Vậy \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le xyz\)
Dấu " = " \(\Leftrightarrow x=y=z\)
Chúc bạn học tốt !!
\(\frac{1+\frac{1}{2}.2.\sqrt{1+x^2}}{x}\le\frac{1+\frac{1}{4}\left(x^2+5\right)}{x}=\frac{x}{4}+\frac{9}{4x}\)
\(\Rightarrow VT\le\frac{1}{4}\left(x+y+z\right)+\frac{9}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(VT\le\frac{1}{4}\left(x+y+z\right)+\frac{9\left(xy+yz+zx\right)}{4xyz}=\frac{1}{4}\left(x+y+z\right)+\frac{9\left(xy+yz+zx\right)}{4\left(x+y+z\right)}\)
\(VT\le\frac{1}{4}\left(x+y+z\right)+\frac{3\left(x+y+z\right)^2}{4\left(x+y+z\right)}=x+y+z=xyz\)
Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)
Ta có :
\(\frac{1+\sqrt{1+x^2}}{x}=\frac{2+\sqrt{4\left(1+x^2\right)}}{2x}\le\frac{2+\frac{4+1+x^2}{2}}{2x}=\frac{9+x^2}{4x}\)
tương tự : \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{9+y^2}{4y}\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{9+z^2}{4z}\)
\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le\frac{\left(9+x^2\right)yz+\left(9+y^2\right)xz+\left(9+z^2\right)xy}{4xyz}\)
\(=\frac{9\left(xy+yz+xz\right)+xyz\left(x+y+z\right)}{4xyz}\le\frac{9\frac{\left(x+y+z\right)^2}{3}+\left(xyz\right)^2}{4xyz}=\frac{4\left(xyz\right)^2}{4xyz}=xyz\)
Dấu " = " xảy ra khi x = y = z = \(\sqrt{3}\)
Đặt \(\frac{1}{1+x}=a\);\(\frac{1}{1+y}=b\);\(\frac{1}{1+y}=c\). Lúc đó a + b + c = 1
Ta có: \(a=\frac{1}{1+x}\Rightarrow x=\frac{1-a}{a}=\frac{\left(a+b+c\right)-a}{a}=\frac{b+c}{a}\)(Do a + b + c = 1)
Tương tự ta có: \(y=\frac{c+a}{b};z=\frac{a+b}{c}\)
\(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\frac{3}{2}\sqrt{xyz}\Leftrightarrow\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}+\frac{1}{\sqrt{xy}}\le\frac{3}{2}\)
Ta đi chứng minh \(\sqrt{\frac{ab}{\left(a+c\right)\left(b+c\right)}}+\sqrt{\frac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\frac{ca}{\left(a+b\right)\left(b+c\right)}}\)\(\le\frac{3}{2}\)
\(VT\le\frac{1}{2}\left(\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{a}{a+b}+\frac{c}{b+c}\right)\)
\(=\frac{1}{2}.3=\frac{3}{2}\)*đúng*
Vậy \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\frac{3}{2}\sqrt{xyz}\)
Đẳng thức xảy ra khi x = y = z = 2
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)
\(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}=\dfrac{2a}{\sqrt{ab+bc+ca+a^2}}+\dfrac{b}{\sqrt{ab+bc+ca+b^2}}+\dfrac{c}{\sqrt{ab+bc+ca+c^2}}\)
\(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)
\(P=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{c+a}.\dfrac{c}{2\left(c+b\right)}}\)
\(P\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{c+a}+\dfrac{c}{2\left(c+b\right)}\right)=\dfrac{9}{4}\)
\(P_{max}=\dfrac{9}{4}\) khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\) hay \(\left(x;y;z\right)=\left(\dfrac{\sqrt{15}}{7};\sqrt{15};\sqrt{15}\right)\)
Áp dụng bđt bunhiacopxki, ta có:
\(\left(x^2+\frac{1}{x^2}\right)\left(1+16\right)\ge\left(x+\frac{4}{x}\right)^2\) => \(x^2+\frac{1}{x^2}\ge\frac{\left(x+\frac{4}{x}\right)^2}{17}\)
=> \(\sqrt{x^2+\frac{1}{x^2}}\ge\frac{x+\frac{4}{x}}{\sqrt{17}}=\frac{x}{\sqrt{17}}+\frac{4}{x\sqrt{17}}\)
CMTT: \(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{y}{\sqrt{17}}+\frac{4}{\sqrt{17}y}\)
\(\sqrt{z^2+\frac{1}{z^2}}\ge\frac{z}{\sqrt{17}}+\frac{4}{\sqrt{17}z}\)
=> A \(\ge\frac{x+y+z}{\sqrt{17}}+\frac{4}{\sqrt{17}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{x+y+z}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}\)(bđt: 1/a + 1/b + 1/c > = 9/(a+b+c)
=> A \(\ge\frac{16\left(x+y+z\right)}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}-\frac{15\left(x+y+z\right)}{\sqrt{17}}\)
A \(\ge2\sqrt{\frac{16\left(x+y+z\right)}{\sqrt{17}}\cdot\frac{36}{\sqrt{17}\left(x+y+z\right)}}-\frac{15\cdot\frac{3}{2}}{\sqrt{17}}\)(Bđt cosi + bđt: x + y + z < = 3/2)
A \(\ge\frac{48}{\sqrt{17}}-\frac{45}{2\sqrt{17}}=\frac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra <=> x = y= z = 1/2
Vậy MinA = \(\frac{3\sqrt{17}}{2}\) <=> x = y = z = 1/2
Áp dụng BĐT Bunhiacopxki :
\(\left(x^2+\frac{1}{x^2}\right)\left(1^2+9^2\right)\ge\left(x+\frac{9}{x}\right)^2\)
\(\Leftrightarrow\sqrt{82}\cdot\sqrt{x^2+\frac{1}{x^2}}\ge x+\frac{9}{x}\)
Chứng minh tương tự :
\(\sqrt{82}\cdot\sqrt{y^2+\frac{1}{y^2}}\ge y+\frac{9}{y}\)
\(\sqrt{82}\cdot\sqrt{z^2+\frac{1}{z^2}}\ge z+\frac{9}{z}\)
Cộng theo vế các BĐT :
\(\sqrt{82}\cdot\left(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}+\sqrt{z^2+\frac{1}{z^2}}\right)\ge x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}\)
Lại có :
\(x+y+z+\frac{9}{x}+\frac{9}{y}+\frac{9}{z}=\frac{9}{x}+81x+\frac{9}{y}+81y+\frac{9}{z}+81z-80\cdot\left(x+y+z\right)\)
\(\ge2\sqrt{\frac{81x\cdot9}{x}}+2\sqrt{\frac{81y\cdot9}{y}}+2\sqrt{\frac{81z\cdot9}{z}}-80\cdot1=82\)
Do đó ta có \(\sqrt{82}\cdot A\ge82\)
\(\Leftrightarrow A\ge\sqrt{82}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
Nhân cả 2 vế với xyz bất đẳng thức sẽ thành yz+ xz+xy+yz\(\sqrt{1+x^2}\)+xz\(\sqrt{1+y^2}+xy\sqrt{1+z^2}\le x^2y^2z^2\)
Ta có yz\(\sqrt{1+x^2}=\sqrt{yz}.\sqrt{yz+x^2yz}=\sqrt{yz}.\sqrt{yz+x\left(x+y+z\right)}=\)\(\sqrt{yz}.\sqrt{\left(x+y\right)\left(x+z\right)}\)\(\le\)\(yz+\frac{\left(x+y\right)\left(x+z\right)}{4}\)(2ab\(\le a^2+b^2\))
làm tương tự ta được xz\(\sqrt{1+x^2}\le xz+\frac{\left(x+y\right)\left(y+z\right)}{4};xy\sqrt{1+z^2}\le xy+\frac{\left(y+z\right)\left(z+x\right)}{4}.\)
vế trái \(\le\) 2(xy+yz+zx) + \(\frac{\left(x+y\right)\left(x+z\right)+\left(y+x\right)\left(y+z\right)+\left(z+x\right)\left(z+y\right)}{4}\)\(\le2.\frac{1}{3}.\left(x+y+z\right)^2+\frac{\frac{1}{3}\left(x+y+y+z+z+x\right)^2}{4}=\left(x+y+z\right)^2=x^2y^2z^2.\)
[ (a-b)2 +(b-c)2 +(c-a)2 \(\ge0\)<=>\(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\) áp dụng vào trên)
dấu '=' xảy ra khi x=y=z \(\sqrt{3}\)