2, `a,b,c ∈ N, a+b+c=2021` Tìm GTNN P=$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$
2 câu trả lời
`P=a/(b+c)+b/(c+a)+c/(a+b)`
`P=a/(b+c)+1+b/(c+a)+1+c/(a+b)+1-3`
`P=(a+b+c)/(b+c)+(a+b+c)/(c+a)+(a+b+c)/(a+b)-3`
`P=(a+b+c)(1/(b+c)+1/(c+a)+1/(a+b))-3`
`P=`$\dfrac{(2a+2b+2c)(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b})}{2}$`-3>=9/2-3=3/2`
Dấu $"="$ xảy ra
`<=>{(a=b=c),(a+b+c=2021):}`
`<=>a=b=c=2021/3`
Vậy `minP=3/2` khi `a=b=c=2021/3`
`P+3=(a+b+c)/(b+c)+(a+b+c)/(a+c)+(a+b+c)/(a+b)`
`=(a+b+c)(1/(b+c)+1/(a+c)+1/(a+b))`
`= 1/2 . (a+b + b+c+a+c)(1/(b+c)+1/(a+c)+1/(a+b))`
Đặt `x=a+b,y=b+c,z=a+c->x,y,z>0`
`->P+3=1/2 (x+y+z)(1/x+1/y+1/z)`
Áp dụng BĐT Cô-si ta có :
$x+y+z\ge 3\sqrt[3]{xyz}\\\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge 3\sqrt[3]{\dfrac{1}{xyz}}$
`-> P+3>=1/2 .9 =9/2`
`-> P>=3/2`
Dấu "`=`" xảy ra khi : `x=y=z`
`<=>a=b=c` mà `a+b+c=2021`
`<=>a=b=c=2021/3`
Vậy `min P=3/2<=>a=b=c=2021/3`