We have:

\(A=\Sigma_{cyc}\frac{1}{3xy+3zx+x+y+z}\le\frac{1}{3xy+3zx+3\sqrt[3]{xyz}}=\Sigma_{cyc}\frac{1}{3xy+3zx+3}=\Sigma_{cyc}\frac{1}{3\left(xy+zx+1\right)}\)

Dat \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)=\left(a;b;c\right)\Rightarrow abc=1\)

\(\Rightarrow A\le\Sigma_{cyc}\frac{1}{3\left(\frac{1}{ab}+\frac{1}{ca}+1\right)}=\Sigma_{cyc}\frac{a}{3\left(a+b+c\right)}=\frac{1}{3}\)

Dau '=' xay ra khi \(x=y=z=1\)

Áp dụng bất đẳng thức Cô-si, ta có: \(\left(3x+1\right)\left(y+z\right)+x=3xy+3xz+\left(x+y+z\right)\ge3xy+3xz+3\sqrt[3]{xyz}\)\(=3xy+3xz+3\Rightarrow\frac{1}{\left(3x+1\right)\left(y+z\right)+x}\le\frac{1}{3\left(xy+xz+1\right)}\)

Tiếp tục áp dụng bất đẳng thức dạng \(u^3+v^3\ge uv\left(u+v\right)\), ta được: \(\frac{1}{3\left(xy+xz+1\right)}=\frac{1}{3\left[x\left(\left(\sqrt[3]{y}\right)^3+\left(\sqrt[3]{z}\right)^3\right)+1\right]}\le\frac{1}{3\left[x\sqrt[3]{yz}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+1\right]}\)\(=\frac{\sqrt[3]{xyz}}{3\left[\sqrt[3]{x^2}\left(\sqrt[3]{y}+\sqrt[3]{z}\right)+\sqrt[3]{xyz}\right]}=\frac{\sqrt[3]{yz}}{3\left(\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}\right)}\)

Tương tự rồi cộng lại theo vế, ta được: \(P\le\frac{1}{3}\)

Đẳng thức xảy ra khi x = y = z = 1

Theo BĐT AM - GM cho 3 số dương, ta có: \(\left(3x+1\right)\left(y+z\right)+x=3xy+3zx+x+y+z\)

\(\ge3xy+3zx+3\sqrt[3]{xyz}=3zx+3xy+3=3\left(zx+xy+1\right)\)(Do xyz = 1)

\(\Rightarrow\frac{1}{\left(3x+1\right)\left(y+z\right)+x}\le\frac{1}{3\left(zx+xy+1\right)}\)(1)

Tương tự ta có: \(\frac{1}{\left(3y+1\right)\left(z+x\right)+y}\le\frac{1}{3\left(xy+yz+1\right)}\)(2); \(\frac{1}{\left(3z+1\right)\left(x+y\right)+z}\le\frac{1}{3\left(yz+zx+1\right)}\)(3)

Cộng theo từng vế của 3 BĐT (1), (2), (3), ta được:  \(P\le\frac{1}{3}\left(\frac{1}{xy+yz+1}+\frac{1}{yz+zx+1}+\frac{1}{zx+xy+1}\right)\)

Ta có BĐT: \(a^3+b^3\ge ab\left(a+b\right)\)

Thật vậy, với a, b dương thì (*)\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\Leftrightarrow a^2-ab+b^2\ge ab\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(đúng)

Áp dụng BĐT trên và sử dụng giả thiết xyz = 1, ta được: \(\frac{1}{xy+yz+1}=\frac{\sqrt[3]{xyz}}{y\left(z+x\right)+\sqrt[3]{xyz}}\)

\(=\frac{\sqrt[3]{xyz}}{y\left[\left(\sqrt[3]{z}\right)^3+\left(\sqrt[3]{x}\right)^3\right]+\sqrt[3]{xyz}}\le\frac{\sqrt[3]{xyz}}{y\sqrt[3]{zx}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{xyz}}\)

\(=\frac{\sqrt[3]{xyz}}{\sqrt[3]{y^3zx}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{xyz}}=\frac{\sqrt[3]{xyz}}{\sqrt[3]{y^2}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{xyz}}\)

\(=\frac{\sqrt[3]{zx}}{\sqrt[3]{y}\left(\sqrt[3]{z}+\sqrt[3]{x}\right)+\sqrt[3]{zx}}=\frac{\sqrt[3]{zx}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}\)(*)

Tương tự: \(\frac{1}{yz+zx+1}\le\frac{\sqrt[3]{xy}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}\)(**); \(\frac{1}{zx+xy+1}\le\frac{\sqrt[3]{yz}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}\)(***)

Cộng theo từng vế của 3 BĐT (*), (**), (***), ta được: \(\frac{1}{xy+yz+1}+\frac{1}{yz+zx+1}+\frac{1}{zx+xy+1}\le\frac{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}{\sqrt[3]{xy}+\sqrt[3]{yz}+\sqrt[3]{zx}}=1\)

\(\Rightarrow P\le\frac{1}{3}\left(\frac{1}{xy+yz+1}+\frac{1}{yz+zx+1}+\frac{1}{zx+xy+1}\right)\le\frac{1}{3}\)

Đẳng thức xảy ra khi x = y = z = 1

https://h.vn//hoi-dap/question/873191.html

https://h.vn/hoi-dap/question/873191.html

Ta có: \(\frac{1}{\left(3x+1\right)\left(y+z\right)+x}=\frac{1}{3x\left(y+z\right)+x+y+z}\le\frac{1}{3x\left(y+z\right)+3\sqrt[3]{xyz}}\)

\(=\frac{1}{3x\left(y+z\right)+3\sqrt[3]{1}}=\frac{1}{3x\left(y+z\right)+3}=\frac{1}{3\left(xy+zx+1\right)}=\frac{1}{3}\cdot\frac{1}{\frac{1}{y}+\frac{1}{z}+1}\)

Tương tự ta chứng minh được:

\(\frac{1}{\left(3y+1\right)\left(z+x\right)+y}\le\frac{1}{3}\cdot\frac{1}{\frac{1}{z}+\frac{1}{x}+1}\) ; \(\frac{1}{\left(3z+1\right)\left(x+y\right)+z}\le\frac{1}{3}\cdot\frac{1}{\frac{1}{x}+\frac{1}{y}+1}\)

Cộng vế 3 BĐT trên lại:

\(A\le\frac{1}{3}\cdot\left(\frac{1}{\frac{1}{x}+\frac{1}{y}+1}+\frac{1}{\frac{1}{y}+\frac{1}{z}+1}+\frac{1}{\frac{1}{z}+\frac{1}{x}+1}\right)\)

\(\Leftrightarrow3A\le\frac{1}{\left(\frac{1}{\sqrt[3]{x}}\right)^3+\left(\frac{1}{\sqrt[3]{y}}\right)^3+1}+\frac{1}{\left(\frac{1}{\sqrt[3]{y}}\right)^3+\left(\frac{1}{\sqrt[3]{z}}\right)^3+1}+\frac{1}{\left(\frac{1}{\sqrt[3]{z}}\right)^3+\left(\frac{1}{\sqrt[3]{x}}\right)^3+1}\)

Đặt \(\left(\frac{1}{\sqrt[3]{x}};\frac{1}{\sqrt[3]{y}};\frac{1}{\sqrt[3]{z}}\right)=\left(a;b;c\right)\) khi đó:

\(3A\le\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}+\frac{1}{c^3+a^3+1}\)

\(=\frac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)+1}+\frac{1}{\left(b+c\right)\left(b^2-bc+c^2\right)+1}+\frac{1}{\left(c+a\right)\left(c^2-ca+a^2\right)+1}\)

\(\le\frac{1}{\left(a+b\right)\left(2ab-ab\right)+1}+\frac{1}{\left(b+c\right)\left(2bc-bc\right)+1}+\frac{1}{\left(c+a\right)\left(2ca-ca\right)+1}\)

\(=\frac{1}{ab\left(a+b\right)+1}+\frac{1}{bc\left(b+c\right)+1}+\frac{1}{ca\left(c+a\right)+1}\)

\(=\frac{abc}{ab\left(a+b\right)+abc}+\frac{abc}{bc\left(b+c\right)+abc}+\frac{abc}{ca\left(c+a\right)+abc}\)

\(=\frac{c}{a+b+c}+\frac{a}{b+c+a}+\frac{b}{c+a+b}\)

\(=\frac{a+b+c}{a+b+c}=1\)

Dấu "=" xảy ra khi: \(a=b=c\Leftrightarrow x=y=z=1\)

Vậy Max(A) = 1 khi x = y = z = 1

Câu hỏi của Pham Van Hung - Toán lớp 9 - Học toán với OnlineMath

\(E= {\sum {(yz)^2 \over xy+zx}}\)>=3/2 (AD BĐT Nesbit)

Dấu = xảy ra <=>x=y=z=1

đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow abc=\frac{1}{xyz}=1\)

Ta có : \(x+y=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}=c\left(a+b\right)\)

Tương tự : \(y+z=a\left(b+c\right);x+z=b\left(c+a\right)\)

\(\Rightarrow E=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{3\sqrt[3]{abc}}{2}=\frac{3}{2}\)

\(\Rightarrow E\ge\frac{3}{2}\)

Vậy GTNN của E là \(\frac{3}{2}\Leftrightarrow x=y=z=1\)

1 cách khác Engel nữa,

\(E=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\)

Áp dụng BĐT Bu-nhi-a-cốp-ski,ta có :

\(\left(a+b+c\right)^2=\left(\frac{a}{\sqrt{b+c}}.\sqrt{b+c}+\frac{b}{\sqrt{c+a}}.\sqrt{c+a}+\frac{c}{\sqrt{a+b}}.\sqrt{a+b}\right)^2\)

\(\le\left(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\right)\left(2a+2b+2c\right)\)

\(\Rightarrow E\ge\frac{a+b+c}{2}\ge\frac{3\sqrt[3]{abc}}{2}=\frac{3}{2}\)

Vậy ....

Lời giải:

Sử dụng bổ đề: Với \(a,b>0\Rightarrow a^3+b^3\geq ab(a+b)\)

BĐT đúng vì nó tương đương với \((a-b)^2(a+b)\geq 0\) (luôn đúng)

Áp dụng vào bài toán:

\(P\leq \frac{1}{x^3yz(y+z)+1}+\frac{1}{y^3xz(x+z)+1}+\frac{1}{z^3xy(x+y)+1}\)

\(\Leftrightarrow P\leq \frac{1}{x^2(y+z)+xyz}+\frac{1}{y^2(x+z)+xyz}+\frac{1}{z^2(x+y)+xyz}\)

\(\Leftrightarrow P\leq \frac{1}{x(xy+yz+xz)}+\frac{1}{y(xy+yz+xz)}+\frac{1}{z(xy+yz+xz)}=\frac{xy+yz+xz}{xy+yz+xz}=1\)

Vậy \(P_{\max}=1\Leftrightarrow x=y=z=1\)

\(P=\dfrac{1}{3x\left(y+z\right)+x+y+z}+\dfrac{1}{3y\left(z+x\right)+x+y+z}+\dfrac{1}{3z\left(x+y\right)+x+y+z}\)

\(P\le\dfrac{1}{3x\left(y+z\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3y\left(z+x\right)+3\sqrt[3]{xyz}}+\dfrac{1}{3z\left(x+y\right)+3\sqrt[3]{xyz}}\)

\(P\le\dfrac{1}{3x\left(y+z\right)+3}+\dfrac{1}{3y\left(z+x\right)+3}+\dfrac{1}{3z\left(x+y\right)+3}\)

Đặt \(\left(x;y;z\right)=\left(a^3;b^3;c^3\right)\Rightarrow abc=1\)

\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3\left(b^3+c^3\right)+1}+\dfrac{1}{b^3\left(c^3+a^3\right)+1}+\dfrac{1}{c^3\left(a^3+b^3\right)+1}\right)\)

\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{1}{a^3bc\left(b+c\right)+1}+\dfrac{1}{b^3ac\left(a+c\right)+1}+\dfrac{1}{c^3ab\left(a+b\right)+1}\right)\)

\(\Rightarrow P\le\dfrac{1}{3}\left(\dfrac{bc}{a\left(b+c\right)+bc}+\dfrac{ac}{b\left(a+c\right)+ac}+\dfrac{ab}{c\left(a+b\right)+ab}\right)=\dfrac{1}{3}\)

\(P_{max}=\dfrac{1}{3}\) khi \(a=b=c=1\) hay \(x=y=z=1\)