Áp dụng BĐT AM-GM:

\(VT=\sum\dfrac{\sqrt{\left(x+y\right)^2-xy}}{4yz+1}\ge\sum\dfrac{\sqrt{\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2}}{\left(y+z\right)^2+1}=\sum\dfrac{\dfrac{\sqrt{3}}{2}\left(x+y\right)}{\left(y+z\right)^2+1}\)

Set \(\left\{{}\begin{matrix}x+y=a\\y+z=b\\z+x=c\end{matrix}\right.\)thì giả thiết trở thành \(a+b+c=3\) và cần chứng minh \(\dfrac{\sqrt{3}}{2}.\sum\dfrac{a}{b^2+1}\ge\dfrac{3\sqrt{3}}{4}\)

\(\Leftrightarrow\sum\dfrac{a}{b^2+1}\ge\dfrac{3}{2}\)( đến đây quen thuộc rồi)

Ta có:\(\sum\dfrac{a}{b^2+1}=\sum a-\sum\dfrac{ab^2}{b^2+1}\ge3-\sum\dfrac{ab^2}{2b}\)(AM-GM)

\(VT\ge3-\sum\dfrac{ab}{2}\ge3-\dfrac{\dfrac{1}{3}\left(a+b+c\right)^2}{2}=\dfrac{3}{2}\)( AM-GM)

Vậy ta có đpcm.Dấu = xảy ra khi a=b=c=1 hay \(x=y=z=\dfrac{1}{2}\)

cảm ơn bạn nhé

Ta có : Áp dụng BĐT Cauchy ba số ở mẫu ta được

\(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\dfrac{x}{\dfrac{y+z+1}{3}}+\dfrac{y}{\dfrac{x+z+1}{3}}+\dfrac{z}{\dfrac{x+y+1}{3}}=\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\)Thấy: \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}\left(?!\right)\)

Ta phải chứng minh:

\(\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{3}\)

\(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{9}\)

Mà \(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}=\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\)

Theo C.B.S

\(\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)

Phải chứng minh

\(\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{9}\)

\(\Leftrightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)

Ta có : \(xy+yz+xz\le x^2+y^2+z^2=3\)

Theo C.B.S : \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)

\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le9\)

\(\Rightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)

=> ĐPCM

\(\dfrac{\sqrt{1\left(x-1\right)}}{x}\le\dfrac{1+x-1}{2x}=\dfrac{1}{2}\) ( cauchy )

TT,\(\dfrac{\sqrt{y-2}}{y}\le\dfrac{1}{2\sqrt{2}};\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\)

cộng vế theo vế => đpcm

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Tham khảo tại đây:

Câu hỏi của Hồ Minh Phi - Toán lớp 9 | Học trực tuyến

Lời giải:

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+xz=xyz\)

\(\Rightarrow x^2+xy+yz+xz=x^2+xyz=x(x+yz)\)

\(\Leftrightarrow x+yz=\frac{x^2+xy+yz+xz}{x}=\frac{(x+y)(x+z)}{x}\)

\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\)

Áp dụng BĐT Bunhiacopxky:\((x+y)(x+z)\geq (x+\sqrt{yz})^2\)

\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}\)

Hoàn toàn tương tự:

\(\sqrt{y+xz}\geq \frac{y+\sqrt{xz}}{\sqrt{y}}\); \(\sqrt{z+xy}\geq \frac{z+\sqrt{xy}}{\sqrt{z}}\)

Cộng theo vế các BĐT đã thu được ta có:

\(\text{VT}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}+\frac{y+\sqrt{xz}}{\sqrt{y}}+\frac{z+\sqrt{xy}}{\sqrt{z}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)

\(\Leftrightarrow \text{VT}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xyz}{\sqrt{xyz}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}=\text{VP}\)

Do đó ta có đpcm.

Dấu bằng xảy ra khi \(x=y=z=3\)

Lời giải:

Để cho gọn đặt \((\sqrt{x}; \sqrt{y}; \sqrt{z})=(a,b,c)\) với \(a,b,c>0\)

Khi đó:

\(A=\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}\)

\(=\frac{1}{2}(\frac{2bc}{a^2+2bc}+\frac{2ac}{b^2+2ac}+\frac{2ab}{c^2+2ab})\)

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

\(=\frac{3}{2}-\frac{1}{2}\underbrace{\left(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\right)}_{M}\)

Áp dụng BĐT Cauchy-Schwarz:

\(M\geq \frac{(a+b+c)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{(a+b+c)^2}{(a+b+c)^2}=1\)

\(\Rightarrow A=\frac{3}{2}-\frac{1}{2}M\leq \frac{3}{2}-\frac{1}{2}=1\)

Vậy \(A_{\max}=1\Leftrightarrow a=b=c\Leftrightarrow x=y=z\)

\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)

\(\Rightarrow xyz\le1\)

\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)

Ta co:

\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)

\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)

\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)

\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Rightarrow A\ge xy+yz+zx\)

Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)

\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))

Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)

\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)

\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)

\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)

\(\ge xy+yz+zx\)

Đẳng thức xảy ra khi x = y = z = 1

thay 1=x+y+z vào nhá , ví dụ x=x(x+y+z) rồi phân tích đa thức thành nhân tử!

thay 1=x+y+z vào nhá , ví dụ x=x(x+y+z) rồi phân tích đa thức thành nhân tử!