Có:

\(\frac{x}{\sqrt{y}}+\sqrt{y}\ge2\sqrt{x};\frac{y}{\sqrt{x}}+\sqrt{x}\ge2\sqrt{y}\)

Cộng theo vế suy ra: \(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}+\sqrt{x}+\sqrt{y}\ge2\sqrt{x}+2\sqrt{y}\)

\(\Rightarrow\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\ge0\)

Đẳng thức xảy ra khi x = y

À quên sửa đề là Gtnn

Gọi \(T=...\)

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

\(T+3=\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\left(\frac{1}{\sqrt{x}+\sqrt{y}}+\frac{1}{\sqrt{y}+\sqrt{z}}+\frac{1}{\sqrt{z}+\sqrt{x}}\right)\)

\(\ge\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right).\frac{\left(1+1+1\right)^2}{2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}=\frac{9}{2}\)\(\Rightarrow\)\(T\ge\frac{9}{2}-3=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)

... 

Đặt \(\hept{\begin{cases}\sqrt{x}=a\\\sqrt{y}=b\\\sqrt{z}=c\end{cases}\left(a,b,c>0\right)}\)

Đặt \(P=\frac{\sqrt{x}}{\sqrt{y}+\sqrt{z}}+\frac{\sqrt{y}}{\sqrt{z}+\sqrt{x}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\)

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

\(\Rightarrow P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)

\(P+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)

\(P+3=\frac{a}{b+c}+\frac{b+c}{b+c}+\frac{b}{c+a}+\frac{c+a}{c+a}+\frac{c}{a+b}+\frac{a+b}{a+b}\)

\(2\left(P+3\right)=2.\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(2\left(P+3\right)=\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

Áp dụng BĐT AM-GM ta có:

\(2\left(P+3\right)\ge3.\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.3.\sqrt[3]{\frac{1}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}}=9.\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.\frac{1}{\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\)

\(\left(\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ne0\right)\)

\(\Leftrightarrow P+3\ge4,5\)

\(\Leftrightarrow P\ge1,5\)

\(P=1,5\Leftrightarrow a=b=c\Leftrightarrow\sqrt{x}=\sqrt{y}=\sqrt{z}\Leftrightarrow x=y=z\)

Vậy \(P_{min}=1,5\Leftrightarrow x=y=z\)

C2: Đặt \(\hept{\begin{cases}\sqrt{x}=a\\\sqrt{y}=b\\\sqrt{z}=c\end{cases}}\)

Đặt \(P=\frac{\sqrt{x}}{\sqrt{y}+\sqrt{z}}+\frac{\sqrt{y}}{\sqrt{x}+\sqrt{z}}+\frac{\sqrt{z}}{\sqrt{x}+\sqrt{y}}\)

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

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

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

Áp dụng BĐT Cauchy-schwarz ta có:

\(P\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)

B tự c/m BĐT phụ \(ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)

\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a+b+c\right)^2}{2.\frac{1}{3}\left(a+b+c\right)^2}=\frac{1}{\frac{2}{3}}=\frac{3}{2}=1,5\)

\(P=1,5\Leftrightarrow a=b=c\Leftrightarrow\sqrt{x}=\sqrt{y}=\sqrt{z}\Leftrightarrow x=y=z\)

Vậy \(P_{min}=1,5\Leftrightarrow x=y=z\)

chịu thua vô điều kiện xin lỗi nha : v

muốn biết câu trả lời lo mà sệt trên google ấy đừng có mà dis:v

\(A=\left[\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right).\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{1}{x}+\frac{1}{y}\right]:\frac{\sqrt{x^3}+y.\sqrt{x}+x\sqrt{y}+\sqrt{y^3}}{\sqrt{x^3y}+\sqrt{xy^3}}\)

\(\Leftrightarrow A=\left[\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}.\frac{2}{\sqrt{x}+\sqrt{y}}+\frac{x+y}{xy}\right]:\frac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{xy}\left(x+y\right)}\)

\(\Leftrightarrow A=\frac{2\sqrt{xy}+x+y}{xy}:\frac{\left(\sqrt{x}+\sqrt{y}\right)^3}{\sqrt{xy}\left(x+y\right)}\)

\(\Leftrightarrow A=\frac{\sqrt{xy}\left(x+y\right)}{xy\left(\sqrt{x}+\sqrt{y}\right)}\)

\(\Leftrightarrow A=\frac{\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\)

sai sót chỗ nào chỉ cho mk nhé. ý kia chốc nx làm nốt

\(A=\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\ge\frac{2}{x+1}+\frac{2}{y+1}+\frac{2}{z+1}\ge\frac{18}{x+y+z+3}=3\)

cảm ơn nha

=3 ban nhe.kn voi minh nha