Ta có: \(3\sqrt{x+2y-1}=\sqrt{9\left(x+2y-1\right)}\le\frac{9+x+2y-1}{2}\)

\(=\frac{x+2y}{2}+4\Leftrightarrow3\sqrt{x+2y-1}-4\le\frac{x+2y}{2}\)(1)

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

Cộng theo vế của 3 BĐT (1), (2), (3), ta được:

\(T=\frac{x}{3\sqrt{x+2y-1}-4}+\frac{y}{3\sqrt{y+2z-1}-4}+\frac{z}{3\sqrt{z+2x-1}-4}\)

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

\(\ge2.\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}=2.\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=2\)(Theo BĐT Bunhiacopxki dạng phân thức)

Đẳng thức xảy ra khi \(x=y=z=\frac{10}{3}\)

ai đó trả lời câu hỏi này đi

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Tự chế

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 \(P_{min}=1,5\Leftrightarrow x=y=z\)

\(\frac{x}{\sqrt{y+z-4}}=\frac{2x}{2\sqrt{y+z-4}}\ge\frac{2x}{\frac{4+y+z-4}{2}}=\frac{4x}{y+z}\)

Tương tự và cộng lại ta có: \(P\ge4\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)\)

\(\Rightarrow P\ge4\left(\frac{x^2}{xz+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\right)\ge\frac{4\left(x+y+z\right)^2}{2\left(xy+xz+yz\right)}\ge\frac{2\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)^2}{3}}=6\)

\(\Rightarrow P_{min}=6\) khi \(x=y=z=4\)

2. Áp dụng bđt \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) :

\(B=\frac{x}{x+x+y+z}+\frac{y}{x+y+y+z}+\frac{z}{x+y+z+z}\) \(=x\cdot\frac{1}{\left(x+y\right)+\left(x+z\right)}+y\cdot\frac{1}{\left(x+y\right)+\left(y+z\right)}+z\cdot\frac{1}{\left(x+z\right)+\left(y+z\right)}\)

\(\le\frac{1}{4}\cdot x\left(\frac{1}{x+y}+\frac{1}{x+z}\right)+\frac{1}{4}y\left(\frac{1}{x+y}+\frac{1}{y+z}\right)+\frac{1}{4}z\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

\(\Rightarrow B\le\frac{1}{4}\left(\frac{x}{x+y}+\frac{y}{x+y}+\frac{y}{y+z}+\frac{z}{y+z}+\frac{x}{x+z}+\frac{z}{x+z}\right)=\frac{3}{4}\)

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

Giải hộ mình với mn

1. Áp dụng bđt Cauchy và bđt quen thuộc \(4ab\le\left(a+b\right)^2\) ta có:

\(D=\frac{ab}{a+b}+\frac{a+b}{4ab}+\frac{3\left(a+b\right)}{4ab}\) \(\ge2\sqrt{\frac{ab}{a+b}\cdot\frac{a+b}{4ab}}+\frac{6}{\left(a+b\right)^2}\)

\(\Rightarrow D\ge1+\frac{3}{2}=\frac{5}{2}\)

Dấu "=" \(\Leftrightarrow a=b=1\)

TA CÓ:

\(P=\frac{4x}{4\sqrt{y+z-4}}+\frac{4y}{4\sqrt{z+x-4}}+\frac{4z}{4\sqrt{x+z-4}}\)

ÁP DỤNG HẰNG ĐẲNG THỨC:

a²+4\(\ge\)4a

\(\Rightarrow P\ge\frac{4x}{y+z-4+4}+\frac{4y}{z+x-4+4}+\frac{4z}{4+z+x-4}=4\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\ge6\)

DẤU BẰNG XẢY RA KHI VÀ CHỈ KHI x=y=z=4

NẾU AI CHƯA HIỂU ĐOẠN 

\(4\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\ge6\)

THÌ LÀM THẾ NÀY NHÉ:
TA CÓ:

\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=\frac{x^2}{x\left(y+z\right)}+\frac{y^2}{y\left(z+x\right)}+\frac{z^2}{z\left(x+y\right)}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\frac{\left(x+y+z\right)^2}{2.\frac{\left(x+y+z\right)^2}{3}}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\)\(\Rightarrow4\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\ge\frac{4.3}{2}=6\)