c/m:\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)(1)
c/m:\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}\ge6.\)(2)
tư(1) và(2) suy ra đpcm
c/m:\(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{3}{2}\)(1)
c/m:\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}\ge6.\)(2)
tư(1) và(2) suy ra đpcm
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Cho a,b,c > 0.Chứng minh rằng
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)\(\frac{2}{a+b}\)+\(\frac{2}{b+c}\)+\(\frac{2}{c+a}\)
b,\(\frac{4}{a}\)+\(\frac{5}{b}\)+\(\frac{3}{c}\)\(\ge\)\(4\left(\frac{3}{a+b}+\frac{2}{b+c}+\frac{1}{c+a}\right)\)
Cho a,b,c >0 thỏa mãn a+b+c\(\le\)\(\frac{3}{2}\).Chứng minh
a,\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\)\(\ge\)6
b,a+ b+ c+ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)\(\ge\)\(\frac{15}{2}\)
Chứng minh : \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{c}+\frac{c}{b}+\frac{b}{a}\)với mọi a; b; c khác 0
Chứng minh rằng với \(a;b;c>0\)thì
\(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
chứng minh rằng : Với a; b; c khác 0 ta có \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)
Cho a,b,c>0 chứng minh \(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\)
Cho a,b,c>0 Chứng minh:
\(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Bài 1 :
Cho a, b, c là 3 cạnh của một tam giác. Chứng minh rằng :
\(\frac{ab}{a+b-c}+\frac{bc}{b+c-a}+\frac{ac}{a+c-b}\ge a+b+c\)
Bài 2 :
Cho a, b, c khác 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Rút gọn : \(Q=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)
Bài 3 :
Chứng minh rằng với mọi a, b, c khác 0 ta luôn có :
\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{c}{b}+\frac{b}{a}+\frac{a}{c}\)