Đặt \(\hept{\begin{cases}b+c+d=x>0\\a+c+d=y>0\\a+b+d=z>0\end{cases}}\)và \(a+b+c=t>0\)

\(\Rightarrow\hept{\begin{cases}a=\frac{y+z+t-2x}{3}\\b=\frac{x+z+t-2y}{3}\\c=\frac{x+y+t-2z}{3}\end{cases}}\)và \(d=\frac{x+y+z-2t}{3}\)

Từ đó ta có:\(Q=\frac{y+z+t-2x}{3x}+\frac{x+z+t-2y}{3y}+\frac{x+y+t-2z}{3z}+\frac{x+y+z-2t}{3t}\)

\(=\frac{y}{3x}+\frac{z}{3x}+\frac{t}{3x}-\frac{2}{3}+\frac{x}{3y}+\frac{z}{3y}+\frac{t}{3y}-\frac{2}{3}+\frac{x}{3z}+\frac{y}{3z}+\frac{t}{3z}-\frac{2}{3}+\frac{x}{3t}+\frac{y}{3t}+\frac{z}{3t}-\frac{2}{3}\)

\(=\left(\frac{y}{3x}+\frac{x}{3y}\right)+\left(\frac{z}{3x}+\frac{x}{3z}\right)+\left(\frac{t}{3x}+\frac{x}{3t}\right)+\left(\frac{z}{3y}+\frac{y}{3z}\right)+\left(\frac{t}{3y}+\frac{y}{3t}\right)+\left(\frac{t}{3z}+\frac{z}{3t}\right)-\frac{8}{3}\)

Áp dụng BĐT AM-GM ta được:

\(\frac{y}{3x}+\frac{x}{3y}\ge2\sqrt{\frac{y}{3x}.\frac{x}{3y}}=\frac{2}{3}\)

CMTT \(\Rightarrow Q\ge\frac{4}{3}\)

Dấu"="xảy ra \(\Leftrightarrow a=b=c=d\)

Đặt \(b+c+d=x;c+d+a=y;a+b+d=z;a+b+c=t\)

Có \(a=\frac{y+z+t-2x}{3}\)

Tương tự :\(b=\frac{x+z+t-2y}{3}\)

\(c=\frac{x+y+t-2z}{3}\)

\(d=\frac{y+x+z-2t}{3}\)

Đặt \(M=\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c}\)

Thay vào biểu thức ta có :

\(M=\frac{\frac{y+z+t-2x}{3}}{x}+\frac{\frac{x+z+t-2y}{3}}{y}+\frac{\frac{x+y+t-2z}{3}}{z}+\frac{\frac{y+x+z-2t}{3}}{t}\)

\(=\frac{1}{3}\left(\frac{y+z+t-2x}{x}+\frac{x+z+t-2y}{y}+\frac{x+y+t-2z}{z}+\frac{x+z+y-2t}{t}\right)\)

\(=\frac{1}{3}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{t}{x}+\frac{x}{t}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)+\left(\frac{t}{y}+\frac{y}{t}\right)+\left(\frac{t}{z}+\frac{z}{t}\right)-8\right]\)

Sử dụng BĐT Cô-si suy ra \(Min_M=\frac{1}{3}.\left(12-8\right)=\frac{4}{3}\)

Dấu bằng xảy ra khi x = y = z = t hay \(b+c+d=a+b+c=c+d+a=b+d+a\) ( tự giải ra a=b=c=d)

Đặt \(N=\frac{b+c+d}{a}+\frac{c+a+d}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d}\)

\(=\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{d}{a}+\frac{a}{d}\right)+\left(\frac{c}{b}+\frac{b}{c}\right)+\left(\frac{d}{c}+\frac{c}{d}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)\)

Sử dụng Cô-si ra \(N\ge12\)

Dấu bằng xảy ra khi a=b=c=d ( tự giải ).

Do đó \(S=M+N\ge\frac{4}{3}+12=13\frac{1}{3}\)

Dấu bằng xảy ra khi \(a=b=c=d\)

\(\)

Áp dụng bđt cô - si cho 2 số không âm, ta được:

\(S=\text{ Σ}_{a,b,c,d}\left(\frac{a}{b+c+d}+\frac{b+c+d}{9a}\right)+\text{ Σ}_{a,b,c,d}\frac{8}{9}.\frac{b+c+d}{9a}\)

\(\ge8\sqrt[8]{\frac{a}{b+c+d}.\frac{b}{c+d+a}.\frac{c}{a+b+d}.\frac{d}{a+b+c}}\)\(\sqrt{\frac{b+c+d}{9a}.\frac{c+d+a}{9b}.\frac{a+b+d}{9c}.\frac{a+b+c}{9d}}\)

\(+\frac{8}{9}\left(\frac{b}{a}+\frac{c}{a}+\frac{d}{a}+\frac{c}{b}+\frac{d}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}+\frac{d}{c}+\frac{a}{d}+\frac{b}{d}+\frac{c}{d}\right)\)

\(\ge\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}\)

Đẳng thức xảy ra khi a = b = c = d

Trl:

Bạn kia trả lời đúng rồi nhoa :))

Hok tốt

~ nhé bạn ~

Ta có: 

\(S=\frac{a-d}{b+d}+\frac{d-b}{c+b}+\frac{b-c}{a+c}+\frac{c-a}{d+a}\)

\(=\left(\frac{a-d}{b+d}+1\right)+\left(\frac{d-b}{c+b}+1\right)+\left(\frac{b-c}{a+c}+1\right)+\left(\frac{c-a}{d+a}+1\right)-4\)

\(=\frac{a+b}{b+d}+\frac{d+c}{c+b}+\frac{b+a}{a+c}+\frac{c+d}{d+a}-4\)

\(=\left(a+b\right)\left(\frac{1}{b+d}+\frac{1}{a+c}\right)+\left(c+d\right)\left(\frac{1}{c+b}+\frac{1}{d+a}\right)-4\)

\(\ge\frac{4\left(a+b\right)}{a+b+c+d}+\frac{4\left(c+d\right)}{a+b+c+d}-4\) (Cauchy Schwars)

\(=\frac{4\left(a+b+c+d\right)}{a+b+c+d}-4=4-4=0\)

Dấu "=" xảy ra khi: a = b = c = d

Vậy Min(S) = 0 khi a = b = c = d

Đúng như mình dự đoán.

 

Cau 2 : Chịu

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\)\(\frac{d}{a+b+c}\)

\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)

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

Mà: \(a+b+c+d\ne0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)

\(\Rightarrow a=b=c=d\)

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

\(\Rightarrow A=1+1+1+1=4\)

số đo slaf

4 

nhe sbn

bài dài 

lắm mình

vhir tiện ghi

thế này thôi

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}\)

\(\Rightarrow1+\frac{a}{b+c+d}=1+\frac{b}{a+c+d}=1+\frac{c}{a+b+d}=1+\frac{d}{a+b+c}\)

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

Mà :\(a+b+c+d=0\Rightarrow b+c+d=a+c+d=a+b+d=a+b+c\)

\(\Rightarrow a=b=c=d\)

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

\(\Rightarrow A=1+1+1+1=4\)

Do a,b,c,d > 0 nên \(b+c+d>0,c+d+a>0,d+a+b>0,a+b+c>0\)

Áp dụng BĐT AM - GM ta có :

\(\frac{a}{b+c+b}+\frac{b+c+d}{a}\ge2\sqrt{\frac{a}{b+c+d}.\frac{b+c+d}{a}}=2\)

Tương tự ta có được điều phải chứng minh

Khi đó \(P\ge2+2+2+2=8\)

b phan d

Theo tính chất của dãy tỉ só bằng nhau ta có:

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

=> a=b=c=d

=> M=1+1+1+1=4

Có: \(\frac{a}{b+c+d}+\frac{b+c+d}{a}=\frac{a}{b+c+d}+\frac{b+c+d}{9a}+\frac{8\left(b+c+d\right)}{9a}\)

\(\ge2\sqrt{\frac{a}{b+c+d}.\frac{b+c+d}{9a}}+\frac{8\left(b+c+d\right)}{9a}\)

\(=\frac{2}{3}+\frac{8\left(b+c+d\right)}{9a}\)

Tương tự ba BĐT còn lại và cộng theo vế thu được:

\(\Sigma_{cyc}\left(\frac{a}{b+c+d}+\frac{b+c+d}{a}\right)=\frac{8}{3}+\frac{8}{9}\left(\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+c}{c}+\frac{a+b+c}{d}\right)\)

\(\ge\frac{8}{3}+\frac{32}{9}\sqrt[4]{\frac{\left(b+c+d\right)\left(c+d+a\right)\left(d+a+c\right)\left(a+b+c\right)}{abcd}}\)

\(\ge\frac{8}{3}+\frac{32}{9}\sqrt[4]{\frac{3^4.abcd}{abcd}}=\frac{40}{3}\)

Đẳng thức xảy ra khi a = b =c = d

P/s: Tính sai chỗ nào tự sửa nhá, dạo này hay nhầm lắm!