Lâu lắm r mới quay lại web :))

Xét : \(2A=\dfrac{2\sqrt{yz}}{x+2\sqrt{yz}}+\dfrac{2\sqrt{xz}}{y+2\sqrt{xz}}+\dfrac{2\sqrt{xy}}{z+2\sqrt{xy}}\)

Áp dụng BĐT AM - GM cho các số dương , ta có :

\(\dfrac{2\sqrt{yz}}{x+2\sqrt{yz}}=\dfrac{x+2\sqrt{yz}-x}{x+2\sqrt{yz}}=1-\dfrac{x}{x+2\sqrt{yz}}\le1-\dfrac{x}{x+x+z}\left(1\right)\)

\(\dfrac{2\sqrt{xz}}{y+2\sqrt{xz}}=\dfrac{y+2\sqrt{xz}-y}{y+2\sqrt{xz}}=1-\dfrac{y}{y+2\sqrt{xz}}\le1-\dfrac{y}{x+y+z}\left(2\right)\)

\(\dfrac{2\sqrt{xy}}{z+2\sqrt{xy}}=\dfrac{z+2\sqrt{xy}-z}{z+2\sqrt{xy}}=1-\dfrac{z}{z+2\sqrt{xy}}\le1-\dfrac{z}{x+y+z}\left(3\right)\)

Cộng từng vế của \(\left(1;2;3\right)\) ta được :

\(2A\le1+1+1-\left(\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{x+y+z}\right)=2\)

\(\Leftrightarrow A\le1\)

Dấu \("="\Leftrightarrow x=y=z\)

\(\Rightarrow A_{Max}=1\Leftrightarrow x=y=z\)

Tham khảo tại đây:

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

\(3-2P=\frac{x}{x+2\sqrt{yz}}+\frac{y}{y+2\sqrt{xz}}+\frac{z}{z+2\sqrt{xy}}\)

\(3-2P\ge\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)

\(\Rightarrow2P\le2\Rightarrow P\le1\)

Dấu "=" xảy ra khi \(x=y=z\)

\(M\le\sqrt{\left(1+1\right)\left(x+y+2\right)}=\sqrt{20}=4\sqrt{5}\)

\(M_{max}=4\sqrt{5}\) khi \(\left\{{}\begin{matrix}x-2=y+4\\x+y=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

Áp dụng BĐT Mincopxki ta có:

\(M=\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+xz+x^2}\)

\(=\sqrt{\left(x+\dfrac{y}{2}\right)^2+\dfrac{3y^2}{4}}+\sqrt{\left(y+\dfrac{z}{2}\right)^2+\dfrac{3z^2}{4}}+\sqrt{\left(z+\dfrac{x}{2}\right)^2+\dfrac{3x^2}{4}}\)

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

\(\ge\sqrt{\left(1+\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}=\sqrt{3}\)

\("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

Lời giải:

Ta có: \(x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\)

Mà \((x-y)^2\geq 0\forall x,y\in\mathbb{R}\Rightarrow x^2+xy+y^2\geq \frac{3}{4}(x+y)^2\)

\(\Rightarrow \sqrt{x^2+xy+y^2}\geq \frac{\sqrt{3}}{2}|x+y|\)

Tương tự:

\(\sqrt{y^2+yz+z^2}\geq \frac{\sqrt{3}}{2}|y+z|; \sqrt{z^2+zx+x^2}\geq \frac{\sqrt{3}}{2}|x+z|\)

Cộng các BĐT trên thu được:

\(M\geq \frac{\sqrt{3}}{2}(|x+y|+|y+z|+|z+x|)\geq \frac{\sqrt{3}}{2}|2x+2y+2z|\)

\(\Leftrightarrow M\geq \frac{\sqrt{3}}{2}.2=\sqrt{3}\)

Vậy \(M_{\min}=\sqrt{3}\Leftrightarrow x=y=z=\frac{1}{3}\)

Ta có \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

\(\Leftrightarrow\left(x-2\sqrt{xy}+y\right)+\left(y-2\sqrt{yz}+z\right)+\left(z-2\sqrt{zx}+x\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2+\left(\sqrt{y}-\sqrt{z}\right)^2+\left(\sqrt{z}-\sqrt{x}\right)^2\ge0\) (luôn đúng)

Vậy \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)

Theo BĐT Cauchy-Schwarz dạng Engel

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

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+y}=\dfrac{y}{y+z}=\dfrac{z}{z+x}\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\end{matrix}\right.\)

\(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

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

Dự đoán dấu bằng có khi (x,y,z)(x,y,z) là các hoán vị (0;1;1).

Từ đó ta đánh giá làm mất căn:

Ta có:

\(4\sqrt{2}.\sqrt{\frac{xy+yz+zx}{x^2+y^2+z^2}}=\frac{8\left(xy+yz+zx\right)}{\sqrt{\left(x^2+y^2+z^2\right).2\left(xy+yz+zx\right)}}\)\(\ge\frac{16\left(xy+yz+zx\right)}{\left(x+y+z\right)^2}\)

Do đó ta chỉ cần có

\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+\frac{16\left(xy+yz+zx\right)}{\left(x+y+z\right)^2}\ge6\)

Không mất tính tổng quát, giả sử \(x\ge y\ge z\) suy ra \(x\ge y>0,z\ge0\)

Khi đó, ta chứng minh BĐT mạnh hơn

\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{16\left(xy+yz+zx\right)}{\left(x+y+z\right)^2}\ge6\)

\(\Leftrightarrow\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}-\frac{8\left(x^2+y^2+z^2\right)}{\left(x+y+z\right)^2}\ge0\)

\(\Leftrightarrow\left(x+y+z\right)^3\left(x+y+2z\right)\ge8\left(x+z\right)\left(y+z\right)\left(x^2+y^2+z^2\right)\)

Hay \(\left(x+y+z\right)^4+z\left(x+y+z\right)^3\ge8z^2\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)\left(x^2+y^2+z^2\right)\)

Theo AM-GM:\(\left(x+y+z\right)^4=\left(x^2+y^2+z^2+2\left(xy+yz+zx\right)\right)^2\ge8\left(xy+yz+zx\right)\left(x^2+y^2+z^2\right)\)

Vậy ta chỉ cần chứng minh \(z\left(x+y+z\right)^3\ge8z^2\left(x^2+y^2+z^2\right)\)

\(BDT\Leftrightarrow\left(x+y+z\right)^3\ge8z\left(x^2+y^2+z^2\right)\)

Ta có:\(\left(x+y+z\right)^3=x^3+y^3+z^3+3x\left(y^2+z^2\right)+3y\left(z^2+x^2\right)+3z\left(x^2+y^2\right)+6xyz\ge x^3+y^3+z^3+3x^2y+3xy^2+5xyz+8z^3+3z\left(x^2+y^2\right)\)

Suy ra \(\left(x+y+z\right)^3-8z\left(x^2+y^2+z^2\right)\ge x^3+y^3+3x^2y+3xy^2+5xyz-5z\left(x^2+y^2\right)\)

\(=x^3+y^3+3x^2y+3xy^2+5z\left(xy-x^2-y^2\right)\ge x^3+y^3+3x^2y+3xy^2+5y\left(xy-x^2-y^2\right)\)

\(\ge x^3+y^3+3x^2y+3xy^2-5y\left(x^2+y^2\right)\)

\(=\left(x^2-y^2+4\right)\left(x-y\right)\ge0\)

BĐT được chứng minh.

v~ để cái này lp 9 thì ko hợp @@

Oh shit!