Áp dụng BĐT cauchy ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\sqrt[3]{abc\cdot\dfrac{1}{abc}}=9\)
Dấu \("="\Leftrightarrow a=b=c\)
Áp dụng BĐT cauchy ta có:
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\sqrt[3]{abc\cdot\dfrac{1}{abc}}=9\)
Dấu \("="\Leftrightarrow a=b=c\)
Cho a , b , c > 0 . CMR : \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Chứng minh bất đẳng thức cô-si với 3 số a,b,c không âm: \(\dfrac{a+b+c}{3}\ge\sqrt[3]{abc}\). Dấu đẳng thức xảy ra khi a=b=c.
Áp dụng chứng minh bất đẳng thức: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
Có thể giúp mình không ạ!
a) \(\dfrac{a^3}{\left(1+b\right)\left(1+c\right)}+\dfrac{b^3}{\left(1+a\right)\left(1+c\right)}+\dfrac{c^3}{\left(1+a\right)\left(1+b\right)}\) biết abc=1
b) \(\dfrac{a^3}{bc}+\dfrac{b^3}{ac}+\dfrac{c^3}{ab}\ge a+b+c\)
c) \(\dfrac{ab}{a^5+ab+b^5}+\dfrac{bc}{b^5+bc+c^5}+\dfrac{ac}{a^5+ac+c^5}\) biết abc=1
Xin cảm ơn các bạn trước ạ!
Cho a , b , c dương
CMR \(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)
cho ba số dương a,b,c .Chứng minh rằng \(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(a+c\right)}+\dfrac{1}{c^2\left(b+a\right)}\ge\dfrac{3}{2}\)
BT1: Cho a,b,c>0. CMR: \(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+\left(c+\dfrac{1}{c}\right)^2>33\)
BT2: Cho a,b,c là các số thực. CMR:
\(a^2+b^2+c^2\ge ab+bc+ac+\dfrac{\left(a-b\right)^2}{26}+\dfrac{\left(b-c\right)^2}{6}+\dfrac{\left(c-a\right)^2}{2009}\)
Mk đang cần gấp. Giúp mk với!!!
\(Choa^3+b^3+c^3=3abc\)
\(a+b+c\ne0\)
Tính \(C=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
bài 2: Cho a,b,ckhác 0
\(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
Tính \(D=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
rút gọn \(\dfrac{1}{2\left(a+b\right)^3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}\right)+\dfrac{3}{2\left(a+b\right)^4}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)+\dfrac{3}{\left(a+b\right)^5}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Cho a,b,c,d>0.Tìm GTNN của
S=\(\left(1+\dfrac{2a}{3b}\right)\left(1+\dfrac{2b}{3c}\right)\left(1+\dfrac{2c}{3d}\right)\left(1+\dfrac{2d}{3a}\right)\)