đề này sai rồi

kết quả = 2044

ÁP DỤNG BĐT Cauchy ta có : 

\(\text{a}_1+\text{a}_2+...+\text{a}_n\ge n^n\sqrt{\text{a}_1.\text{a}_2....\text{a}_n}\)  (1) 

\(\frac{1}{\text{a}_1}+\frac{1}{\text{a}_2}+...+\frac{1}{\text{a}_n}\ge n^n\sqrt{\frac{1}{\text{a}_1}\cdot\frac{1}{\text{a}_2}\cdot...\cdot\frac{1}{\text{a}_n}}\)(2) 

Nhân (1) và (2) vế với vế tương ứng ta có được BĐT (*) 

Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}\text{a}_1=\text{a}_2=...=\text{a}_n\\\frac{1}{\text{a}_1}=\frac{1}{\text{a}_2}=...=\frac{1}{\text{a}_n}\end{cases}}\)

                             \(\Leftrightarrow\text{a}_1=\text{a}_2=...=\text{a}_n\)

Không bt lm . Ahihi!

Ta có: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

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

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

Mà: \(\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)+\)\(\left(\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}\right)=3\)

\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+\)\(\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)\)\(=\frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}\ge3\)( BĐT Cosi cho 3 số )

\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+\)\(\left(\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}\right)\)\(=\frac{a+c}{b+c}+\frac{b+a}{c+a}+\frac{c+b}{a+b}\ge3\)( BĐT Cosi cho 3 số )

\(\Rightarrow2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+\)\(\left(\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\right)+\)\(\left(\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}\right)\ge6\)

\(\Rightarrow\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\left(đpcm\right)\)Đẳng thức trên xảy ra khi\(a=b=c\)

Đặt b+c=x , c+a=y , a+b=z

=> a+b+c=(x+y+z)/2 
=> a=(y+z-x)/2 và b=(x+z-y)/2 và c=(x+y-z)/2 
VT = a/(a+b) +b/(b+c) +c/(c+a) 
=(y+z-x)/(2x) + (x+z-y)\(2y) + (x+y-z)/(2z) 
=(y/x + z/x -1+ x/y + z/y -1+ x/z + y/z -1 )/2 
=( y/x+ z/x + x/y + z/y + x/z + y/z -3 )/2 
Áp dụng BĐT Cô si 
( y/x + x/y ) + (z/y + y/z) + (x/z+ z/x) >= 2x3 =6

<=>( y/x + x/y ) + (z/y + y/z) + (x/z+ z/x) -3 >= 3

<=> [( y/x + x/y ) + (z/y + y/z) + (x/z+ z/x) -3]/2 >= 3/2

<=> VT >= 3/2 
Dấu = xảy ra khi: x=y=z <=> a=b=c

\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\)

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

\(\Leftrightarrow\frac{1}{1+a}=\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\left(\text{ta áp dụng BĐT cô-si}\right)\)

       \(\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+c\right)}}\)

Tương tự, ta có: 

\(\frac{1}{1+c}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+b\right)}}\)

Nhân theo vế. ta có: \(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge8\frac{\sqrt{a^2b^2c^2}}{\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}=\frac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Leftrightarrow abc\le\frac{1}{8}\)

Dấu "=" xảy ra khi: \(Q=abc;MAX_Q=\frac{1}{8}\Leftrightarrow a=b=c=\frac{1}{2}\)

P/s: Ko chắc

Dùng cauchy-schawarz là ra nhé :)

m có bị điên ko :v sao copy luôn cái url mess :v 

\(T=3\sqrt{2}+2\sqrt{3}-\sqrt{3^2.2}+\sqrt{4\left(4-4\sqrt{3}+3\right)}\)

=> \(T=3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+\sqrt{2^2\left(2-\sqrt{3}\right)^2}\)

=> \(T=3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+2\left(2-\sqrt{3}\right)\)

=> \(T=3\sqrt{2}+2\sqrt{3}-3\sqrt{2}+4-2\sqrt{3}\)

=> T=4