a)Áp dụng bđt Cô-si:
\(\dfrac{a}{b}+\dfrac{b}{a}-1+\dfrac{ab}{a^2-ab+b^2}=\dfrac{a^2+b^2-ab}{ab}+\dfrac{ab}{a^2-ab+b^2}\ge2\sqrt{\dfrac{a^2+b^2-ab}{ab}.\dfrac{ab}{a^2-ab+b^2}}=2\)
=>\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{ab}{a^2-ab+b^2}\ge3\)
Dấu "=" xảy ra khi a=b=1
b) bđt sai rồi
CMR với a, b, c > 0 thì :
a) \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)
b)\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\le\dfrac{a+b+c}{2}\)
CMR với a, b, c > 0 thì
a) \(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{a}\)
b) \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge a+b+c\)
c) \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
Cho a, b, c là độ dài 3 cạnh của 1 tam giác. CMR:
a) \(a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
b) \(\dfrac{a}{b+c-a}+\dfrac{b}{a+c-b}+\dfrac{c}{a+b-c}\ge3\)
c) \(\dfrac{1}{a+b},\dfrac{1}{b+c},\dfrac{1}{c+a}\)cũng là độ dài ba cạnh của 1 tam giác
Giải hộ mình mấy câu này với
Chứng minh:
a)\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a+b+c\)
b)\(\dfrac{a^2}{c}+\dfrac{b^2}{a}+\dfrac{c^2}{b}\ge a+b+c\)
Cho a, b, c > 0 .CMR: \(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ac}{a+c}\) ≤ \(\dfrac{1}{2}\)
Cho a, b, c > 0 .CMR: \(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ac}{a+c}\) ≤ \(\dfrac{1}{2}\left(a+b+c\right)\)
1,cho a,b,c. cm 1\(\ge\) a,b,c\(\ge\)0, a2 +b2+c2\(\le\)2
2, cho a,b,c>0.
CM: (a+b+c)(\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\))
3,cho a,b,c>0 CM:
\(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)
mn giải giúp mk nha :)
Cho a,b,c>0 và ab+bc+ca=3
CMR: \(\dfrac{1}{a^2+b^2+1}+\dfrac{1}{b^2+c^2+1}+\dfrac{1}{c^2+a^2+1}\le1\)
Cjo 3 số dương a,b,c . CMR
\(\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}\ge\dfrac{a+b+c}{2}\)
