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

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!

\(=>A=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)

áp dụng BĐT AM-GM

\(=>\sqrt{x-1}\le\dfrac{x-1+1}{2}=\dfrac{x}{2}\)

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

có \(\dfrac{\sqrt{y-2}}{y}=\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\)

\(=>\sqrt{\left(y-2\right)2}\le\dfrac{y-2+2}{2}=\dfrac{y}{2}\)

\(=>\dfrac{\sqrt{\left(y-2\right)2}}{\sqrt{2}.y}\le\dfrac{\dfrac{y}{2}}{\sqrt{2}.y}=\dfrac{1}{2\sqrt{2}}\left(2\right)\)

tương tự \(=>\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\left(3\right)\)

(1)(2)(3)\(=>A\le\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}\)

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

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

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

\(\Rightarrow\dfrac{x}{x+\sqrt{x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Tương tự:

\(\dfrac{y}{y+\sqrt{y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

\(\dfrac{z}{z+\sqrt{z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)

Cộng vế:

\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)

Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{3}\)

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 \("="\Leftrightarrow x=y=z\)

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

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

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)

=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)