\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
=> \(A+3=\frac{a}{b+c}+1+\frac{b}{c+a}+1+\frac{c}{a+b}+1\)
\(=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(=\frac{1}{2}\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\ge\frac{1}{2}.3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}.3\sqrt[3]{\frac{1}{a+b}.\frac{1}{b+c}.\frac{1}{c+a}}=\frac{9}{2}\) (AM - GM)
=> \(A\ge\frac{9}{2}-3=\frac{3}{2}\) (đpcm)
Đặt \(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
\(A=\frac{a^2}{ba+ca}+\frac{b^2}{cb+ba}+\frac{c^2}{ac+bc}\)
Áp dụng BĐT Cauchy-schwarz ta có:
\(A=\frac{a^2}{ba+ca}+\frac{b^2}{cb+ba}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2.\left(ab+bc+ca\right)}\)
Ta c/m BĐT phụ \(ab+bc+ca\le\frac{1}{3}.\left(a+b+c\right)^2\)( tự c/m)
Áp dụng:
\(A\ge\frac{\left(a+b+c\right)^2}{2.\frac{1}{3}\left(a+b+c\right)^2}=\frac{1}{\frac{2}{3}}=\frac{3}{2}\)
đpcm
Tham khảo nhé~
P = a/(b+c) + b/(c+a) + c/(a+b)
P + 3 = 1+ a/(b+c) + 1+ b/(c+a) + 1+ c/(a+b)
P + 3 = (a+b+c)/(b+c) + (a+b+c)/(b+c) + (a+b+c)/(c+a)
P + 3 = (a+b+c)[1/(b+c) + 1/(c+a) + 1/(a+b)] (*)
ad bđt cô si cho 3 số:
2(a+b+c) = (a+b) + (b+c) + (c+a) ≥ 3.³√(a+b)(b+c)(c+a)
1/(b+c) + 1/(c+a) + 1/(a+b) ≥ 3.³√1/(a+b)(b+c)(c+a)
nhân lại vế theo vế 2 bđt: 2(a+b+c)[1/(b+c) + 1/(c+a) + 1/(a+b)] ≥ 9
=> P + 3 ≥ 9/2 => P ≥ 3/2 (đpcm) ; dấu "=" khi a = b = c
- - -
cách khác: P = a/(b+c) + b/(c+a) + c/(a+b)
M = b/(b+c) + c/(c+a) + a/(a+b)
N = c/(b+c) + a/(c+a) + b/(a+b)
Thấy: M + N = 3
P + M = (a+b)/(b+c) + (b+c)/(c+a) + (c+a)/(a+b) ≥ 3 (cô si cho 3 số)
P + N = (a+c)/(b+c) + (b+a)/(c+a) + (c+b)/(a+b) ≥ 3 (cô si)
=> 2P + M + N ≥ 6 => 2P + 3 ≥ 6 => P ≥ 3/2 (đpcm) ; đẳng thức khi a = b = c