Cho a,b,c > 0 CMR :
\(a+b+c+\frac{9abc}{ab+bc+ca}\ge4(\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ca}{c+a})\)
Cho các số thực dương a,b,c . CMR :
\(a+b+c+\frac{9abc}{ab+bc+ca}\ge4\left(\frac{ab}{a+b}+\frac{bc}{c+b}+\frac{ca}{c+a}\right)\)
Cho các số thực dương a,b,c . CMR :
\(a+b+c+\frac{9abc}{ab+bc+ca}\ge4\left(\frac{ab}{a+b}+\frac{bc}{c+b}+\frac{ca}{c+a}\right)\)
Cho các số thực dương a,b,c . CMR :
\(a+b+c+\frac{9abc}{ab+bc+ca}\ge4\left(\frac{ab}{a+b}+\frac{bc}{c+b}+\frac{ca}{c+a}\right)\)
Cho các số thực dương a,b,c . CMR :
\(a+b+c+\frac{9abc}{ab+bc+ca}\ge4\left(\frac{ab}{a+b}+\frac{bc}{c+b}+\frac{ca}{c+a}\right)\)
Cho a,b,c>0. Cmr: a) \(\frac{ab}{a^2+bc+ca}+\frac{bc}{b^2+ca+ab}+\frac{ca}{c^2+ab+bc}\le\frac{a^2+b^2+c^2}{ab+bc+ca}\)
b) \(\frac{a}{a^3+b^2+c}+\frac{b}{b^3+c^2+a}+\frac{c}{c^3+a^2+b}\le1\)
a)\(VT=\sum_{cyc}\frac{ab^3+ab^2c+a^2bc}{\left(a^2+bc+ca\right)\left(b^2+bc+ca\right)}\le\frac{\sum_{cyc}\left(ab^3+ab^2c+a^2bc\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{ab^3+bc^3+ca^3+2a^2bc+2ab^2c+2abc^2}{\left(ab+bc+ca\right)^2}\)\(\le\frac{\sum_{cyc}ab\left(a^2+b^2\right)+abc\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}\)
\(=\frac{\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}{\left(ab+bc+ca\right)^2}=\frac{a^2+b^2+c^2}{ab+bc+ca}=VP\)
Cho a,b,c > 0 . Cmr: \(a^2+b^2+c^2+\frac{9abc}{a+b+c}-2\left(ab+bc+ca\right)\ge0\)
Cho a+b+c=1 (a,b,c>0). CMR: \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{3}{2}\)
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Cho a,b,c>0 và a+b+c=1. CMR: \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{3}{2}\)
Ta có : a + bc = a ( a + b + c ) + bc = ( a + c ) ( a + b )
BĐT cần chứng minh tương đương với :
\(\frac{a\left(a+b+c\right)-bc}{\left(a+c\right)\left(a+b\right)}+\frac{b\left(a+b+c\right)-ca}{\left(b+c\right)\left(b+a\right)}+\frac{c\left(a+b+c\right)-ab}{\left(c+a\right)\left(c+b\right)}\le\frac{3}{2}\)
\(\left(a^2+ab+ac-bc\right)\left(b+c\right)+\left(ab+b^2+bc-ac\right)\left(a+c\right)+\left(ac+bc+c^2-ab\right)\left(a+b\right)\le\frac{3}{2}\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
khai triển ra , ta được :
\(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+6abc\le\frac{3}{2}\left(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\right)+3abc\)
\(\Rightarrow\frac{-1}{2}\left(a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\right)\le-3abc\)
\(\Rightarrow a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\ge6abc\)( nhân với -2 thì đổi dấu )
\(\Rightarrow b\left(a^2-2ac+c^2\right)+a\left(b^2-2bc+c^2\right)+c\left(a^2-2ab+b^2\right)\ge0\)
\(\Rightarrow b\left(a-c\right)^2+a\left(b-c\right)^2+c\left(a-b\right)^2\ge0\)
vì BĐT cuối luôn đúng nên BĐT lúc đầu đúng
Dấu " = " xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)
Cho a;b;c > 0. thỏa mãn \(ab+bc+ca+2abc=1\).
CMR: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4\left(a+b+c\right)\)