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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 \(x=y=z=\sqrt{3}\)
Ta chứng minh
\(a+b\ge\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\)
\(\Leftrightarrow\left(\sqrt[3]{a}-\sqrt[3]{b}\right)^2\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\ge0\)(đúng )
Áp đụng vào bài toán ta được
\(\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\)
\(\le\frac{1}{\sqrt[3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)+1}+\frac{1}{\sqrt[3]{yz}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+1}+\frac{1}{\sqrt[3]{zx}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+1}\)
\(=\frac{\sqrt[3]{z}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}+\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}+\frac{\sqrt[3]{y}}{\sqrt[3]{x}+\sqrt[3]{y}+\sqrt[3]{z}}=1\)
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
Áp dụng BĐT AM-GM ta có: \(\frac{x^2+4yz}{2}\ge2x\sqrt{yz}\)
\(\Rightarrow\frac{2}{x^2+4yz}\le\frac{1}{2x\sqrt{yz}}\Rightarrow\frac{1}{x^2+4yz}\le\frac{1}{4x\sqrt{yz}}\)
Cộng theo vế ta có:
\(\frac{1}{x^2+4yz}+\frac{1}{y^2+4xz}+\frac{1}{z^2+4xy}\le\frac{1}{4x\sqrt{yz}}+\frac{1}{4y\sqrt{xz}}+\frac{1}{4z\sqrt{xy}}\)
Cần chứng minh \(\frac{1}{4x\sqrt{yz}}+\frac{1}{4y\sqrt{xz}}+\frac{1}{4z\sqrt{xy}}\le\frac{1}{xyz}\)
Nhân 2 vế với \(xyz\) ta lại được BĐT cần c/m tương đương với:
\(\frac{1}{4}\left(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\right)\le1\)
Áp dụng BĐT AM-GM lần nữa ta có:
\(\frac{1}{4}\left(\sqrt{yz}+\sqrt{xz}+\sqrt{xy}\right)\le\frac{1}{4}\left(x+y+z\right)=1\) (Đúng)
Vậy BĐT đầu đã được c/m
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 \(\sqrt{3}\)
\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
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 \(\left(x;y;z\right)=\left(\dfrac{\sqrt{15}}{7};\sqrt{15};\sqrt{15}\right)\)