cho a,b,c duong , a+b+c=1
a, tim Min A=1/(a^2+b^2) +1/(b^2+c^2) +1/(c^2+a^2) +1/ab +1/bc +1/ac
b, tìm Min B=1/(a^2+bc) +1/(b^2+ac) +1/(c^2+ab) +1/ab +1/bc +1/ac
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
TA
1
1 tháng 1 2018
Theo C.B.S thì
\(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\)
\(\Rightarrow\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ac}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)
Lại theo CBS thì
\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge\dfrac{9}{\left(a+b+c\right)^2}=9\)mà \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\)
\(\Rightarrow\dfrac{7}{ab+bc+ac}\ge21\)
\(\Rightarrow\)\(\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+bc+ac}\)\(\)\(\ge21+9=30\)
vậy Min = 30 khi a = b = c = 1/3
TN
31 tháng 7 2017
\(P=\frac{1}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(=\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\left(1\right)\)
Áp dụng BĐT AM-GM ta có: :
\(\frac{a}{a^2+b^2+c^2}+9a\left(a^2+b^2+c^2\right)\ge2\sqrt{9a^2}=6a\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{b}{a^2+b^2+c^2}+9b\left(a^2+b^2+c^2\right)\ge6b;\frac{c}{a^2+b^2+c^2}+9c\left(a^2+b^2+c^2\right)\ge6c\)
\(\Rightarrow\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}+9\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge6\left(a+b+c\right)\)
Theo BĐT Cauchy-Schwarz thì:
\(9\left(a^2+b^2+c^2\right)\left(a+b+c\right)\ge9\cdot\frac{\left(a+b+c\right)^2}{3}\cdot\left(a+b+c\right)=3\)
\(\Rightarrow\frac{a}{a^2+b^2+c^2}+\frac{b}{a^2+b^2+c^2}+\frac{c}{a^2+b^2+c^2}\ge6-3=3\)
Và \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{9}{ab+bc+ca}\ge\frac{9}{\frac{\left(a+b+c\right)^2}{3}}=27\)
Khi đó nhìn vào \(\left(1\right)\) thấy \(P\ge27+3=30\)
Xảy ra khi \(a=b=c=\frac{1}{3}\)
M
1
18 tháng 11 2017
\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}+\frac{1}{\left(a-b\right)\left(c^2+bc-a^2-ab\right)}\)
\(=\frac{c-a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}+\frac{a-b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}\)
\(+\frac{b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=0\)
\(a\text{) }\)Áp dụng: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) (a, b > 0). Dấu "=" xảy ra khi a = b.
\(\frac{1}{a^2+b^2}+\frac{1}{ab}=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}\ge\frac{4}{a^2+b^2+2ab}+\frac{1}{2.\frac{\left(a+b\right)^2}{4}}=\frac{6}{\left(a+b\right)^2}\)
\(=6\left[\frac{1}{\left(a+b\right)^2}+\frac{27}{8}\left(a+b\right)+\frac{27}{8}\left(a+b\right)\right]-\frac{81}{2}\left(a+b\right)\)
\(\ge6.3\sqrt[3]{\frac{1}{\left(a+b\right)^2}.\frac{27}{8}\left(a+b\right).\frac{27}{8}\left(a+b\right)}-\frac{81}{2}\left(a+b\right)\)
\(=\frac{81}{2}-\frac{81}{2}\left(a+b\right)\)
Tương tự: \(\frac{1}{b^2+c^2}+\frac{1}{bc}\ge\frac{81}{2}-\frac{81}{2}\left(b+c\right)\)
\(\frac{1}{c^2+a^2}+\frac{1}{ca}\ge\frac{81}{2}-\frac{81}{2}\left(c+a\right)\)
Cộng theo vế ta được
\(A\ge3.\frac{81}{2}-81\left(a+b+c\right)=3.\frac{81}{2}-81=\frac{81}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}.\)
Vậy GTNN của A là \(\frac{81}{2}.\)