Cho a,b,c >0. C/m $\frac{a^{2}}{b+c}$+$\frac{b^{2}}{a+c}$+$\frac{c^{2}}{a+b}$$\geq$ $\frac{a+b+c}{2}$

Đáp án:

Giải thích các bước giải:

 Cách `1`

Chứng minh BĐT phụ

`a_1^2/b_1 +a_2^2/b_2>=(a_1+a_2)^2/(b_1+b_2) (b_1,b_2>0)`

`<=>(a_1^2 b_2 +a_2^2 b_1)/(b_1b_2)>=(a_1+a_2)^2/(b_1+b_2)`

`<=>(a_1^2 b_2 +a_2^2 b_1)(b_1+b_2)>=(a_1+a_2)^2 b_1b_2`

`<=>a_1^2 b_2^2 +a_2^2+b_1^2 +a_1^2 b_1b_2+a_2^2 b_1b_2 >=a_1^2 b_1b_2+a_2^2 b_1b_2 +2a_1a_2b_1b_2`

`<=>(a_1b_2)^2 +(a_2b_1)^2>=2a_1a_2b_1b_2`

`<=>(a_1b_2)^2 -2a_1a_2b_1b_2+(a_2b_1)^2>=0`

`<=>(a_1b_2-a_2b_1)^2>=0`

Dấu `=` xảy ra `<=>a_1/b_1=a_2/b_2`

`=>a_1^2/b_1 +a_2^2/b_2+a_3^2/b_3 >=(a_1+a_2)^2/(b_1+b_2) +a_3^2/b_3 >=(a_1+a_2+a_3)^2/(b_1+b_2+b_3)`

`ĐK:b_1,b_2,b_3>0`

Dấu `=` xảy ra `<=>a_1/b_1=a_2/b_2=a_3/b_3`

Áp dụng: Do `a,b,c>0`

`=>a^2/(b+c)+b^2/(c+a)+c^2/(a+b)>=(a+b+c)^2/[(b+c)+(c+a)+(a+b)]`

`=>a^2/(b+c)+b^2/(c+a)+c^2/(a+b)>=(a+b+c)^2/[2(a+b+c)]`

`=>a^2/(b+c)+b^2/(c+a)+c^2/(a+b)>=(a+b+c)/2`

Dấu `=` xảy ra `<=>a=b=c`

Cách `2`

Do `a,b,c>0` ,Áp dụng BĐT Co-si

`a^2/(b+c) +(b+c)/4 >=2\sqrt{a^2/(b+c) .(b+c)/4} `

`a^2/(b+c) +(b+c)/4 >=2 . a/2`

`a^2/(b+c) +(b+c)/4 >=a`

Tương tự

`b^2/(c+a) +(c+a)/4 >=b; c^2/(a+b) +(a+b)/4 >=c`

`=>a^2/(b+c) +(b+c)/4 +b^2/(c+a) +(c+a)/4 + c^2/(a+b) +(a+b)/4>=a+b+c`

`=>a^2/(b+c)+b^2/(c+a)+c^2/(a+b) +(a+b+c)/2 >=a+b+c`

`=>a^2/(b+c)+b^2/(c+a)+c^2/(a+b)>=(a+b+c)/2`

Dấu `=` xảy ra `<=>a=b=c`