Ta có:
\(\dfrac{bc}{a}+\dfrac{ac}{b}=c\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2c\)
Chứng minh tương tự, ta có:
\(\dfrac{bc}{a}+\dfrac{ab}{c}\ge2b\)
\(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\)
\(\Rightarrow2\left(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)
Dấu = xảy ra khi a = b = c
Cho a,b,c > 0 . CMR :
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
Cho a,b,c >0. Chứng minh rằng: \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)
Cho \(a,b,c>0\) thỏa mãn \(ab+bc+ca=3\) . CMR : \(\sqrt[3]{\dfrac{a}{b\left(b+2c\right)}}+\sqrt[3]{\dfrac{b}{c\left(c+2a\right)}}+\sqrt[3]{\dfrac{c}{a\left(a+2b\right)}\ge\dfrac{3}{\sqrt[3]{3}}}\)
Cho \(a;b;c>0\). CMR:
\(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge ab+bc+ca\)
Cho a,b,c>0 thỏa mãn a+b+c\(\le3\).
CMR : A= \(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{3}{2}\)
(Sử dụng Cauchy)
Cho biết: \(\dfrac{a+b-c}{ab}-\dfrac{b+c-a}{bc}-\dfrac{a+c-b}{ac}=0\). CMR trong 3 phân thức ở vế trái, có ít nhất một phân thức bằng 0
Cho a,b,c>0 thỏa mãn a+b+c\(\le\)3
CMR:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ca}\ge\dfrac{3}{2}\)
(Sử dụng Cauchy)
Cho a>0; b>0; c>0. CMR:
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\ge\dfrac{3}{a+b+c}\)
Bài 1: Cho a,b,c là những số dương thỏa mãn: a+b+c=3
CMR: \(\dfrac{a^2}{a+2b^3}+\dfrac{b^2}{b+2c^3}+\dfrac{c^2}{c+2a^3}\ge1\)
Bài 2: Cho a, b, c thỏa mãn: ab+bc+ca=3
CMR: \(\dfrac{a}{2b^3+1}+\dfrac{b}{2c^3+1}+\dfrac{c}{2a^3+1}\ge1\)
Bài 3: Cho a, b, c > 0. CMR: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+3b\)
Dấu = xảy ra khi a=b=2c
