\(P=\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{b+c}\right)+\dfrac{3a}{a+c}\)
\(=\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{a}{a+c}\right)+\dfrac{1}{2}\left(\dfrac{c}{b+c}+\dfrac{a}{a+c}\right)+\dfrac{2a}{a+c}\)
\(P\ge\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)+\dfrac{1}{2}\left(\dfrac{c}{a+c}+\dfrac{a}{a+c}\right)+\dfrac{2a}{a+a}=2\)
Dấu "=" xảy ra khi \(a=b=c\)
Biết a,b,c > 0 thỏa mãn ab+bc+ca=3abc
\(P=\dfrac{a}{\left(3a-1\right)^2}+\dfrac{b}{\left(3b-1\right)^2}+\dfrac{c}{\left(3c-1\right)^2}\) đạt min
Cho a,b,c > 0 thỏa a+b+c=abc. Tìm GTLN của BT :
\(\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}+\dfrac{b}{\sqrt{ac\left(1+b^2\right)}}+\dfrac{c}{\sqrt{ab\left(1+c^2\right)}}\)
Cho \(a,b,c>0\). Tìm min:
\(P=\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}-\dfrac{12abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
1. Cho a,b >0; a+b ≤ 1
Tìm min \(N=ab+\dfrac{1}{ab}\)
2. Cho a,b,c >0 t/m: a+b+c ≥ 6
Tìm min \(P=5a+6b+7c+\dfrac{1}{a}+\dfrac{8}{b}+\dfrac{27}{c}\)
3. Cho a,b,c ∈ \(\left[-1;2\right]\) và \(a^2+b^2+c^2=6\)
\(CM:\) a+b+c ≥ 0
Cho a, b, c là các số thực dương thỏa mãn b2 + c2 ≤ a2. Tìm Min:\(M=\dfrac{1}{a^2}\left(b^2+c^2\right)+a^2\left(\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\)
Cho a, b, c khác 0 và \(a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
Tính \(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
Cho a,b,c khác 0 thỏa mãn a\(\left(\dfrac{1}{c}+\dfrac{1}{b}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-2\)
a(1b+1c)+b(1c+1a)+c(1a+1b)=−2
và a3+b3+c3=1. CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Cho a,b,c khác 0 thỏa mãn \(a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-2\)
và a3+b3+c3=1. CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
a,b,c là các số thực dương. Tìm Min \(P=\dfrac{2a^2+ab}{\left(b+\sqrt{ca}+c\right)^2}+\dfrac{2b^2+bc}{\left(c+\sqrt{ab}+a\right)^2}+\dfrac{2c^2+ca}{\left(a+\sqrt{bc}+b\right)^2}\)
cho a,b,c>0
CMR:
1) \(a+b+\dfrac{1}{4}\ge\sqrt{a+b}\)
2) \(\left(a+b+\dfrac{1}{2}\right)^2+\left(b+c+\dfrac{1}{2}\right)^2+\left(c+a+\dfrac{1}{2}\right)^2\ge4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\right)\)
