Cho `a,b,c>0`. CMR: (câu nào cx đc) a) `(a+b)/(a+b+2c)+(b+c)/(b+c+2a)+(c+a)/(c+a+2b)\le 3/2` b) `a/(3a+b+c)+b/(3b+c+a)+c/(3c+b+a)\le 3/5`

Đáp án+Giải thích các bước giải:

Gọi `P=a/(3a+b+c)+b/(3b+c+a)+c/(3c+b+a)\le 3/5`

Đặt `3a+b+c=x` `=>x-2a=b+c`

       `3b+c+a=y` `=>y-2b=c+a`

       `3c+a+b=z` `=>z-2c=a+b`

Cộng vế, ta có:

`x+y+z=5(a+b+c)=5(x-2a)=5(y-2b)=5(z-2c)`

`=>`$\left\{\begin{matrix}x+y+z=5(x-2a) \\ x+y+z=5(y-2b) \\ x+y+z=5(z-2c) \end{matrix}\right.$

`<=>`$\left\{\begin{matrix}4x-y-z=10a \\ 4y-x-z=10b \\ 4z-y-x=10c \end{matrix}\right.$

$\left\{\begin{matrix}\frac{(4x-y-z)}{x}= \frac{10a}{x}
\\ \frac{(4y-x-z)}{y}= \frac{10b}{y}
\\ \frac{(4z-y-x)}{z}= \frac{10c}{z} 
\end{matrix}\right.$

Lại cộng vế, ta có:

`10P=(4x-(y+z))/x+(4y-(z+x))/y+(4z-(x+y))/z`

`<=>10P=(4x)/x-(y+z)/x+(4y)/y-(z+x)/y+(4z)/z-(x+y)/z`

`<=>10P=4-y/x-z/x+4-z/y-x/y+4-x/z-y/z`

`<=>10P=12-(y/x+z/x+z/y+x/y+x/z+y/z)`

Mà `(y/x+z/x+z/y+x/y+x/z+y/z)>=6`

`=>-(y/x+z/x+z/y+x/y+x/z+y/z)<=6`

`=>10P<=12-6=6`

`=>P<=3/5`

Dấu "=" xảy ra `<=>` `a=b=c`