Cho 3 ssos dương a,b,c có a+b+c=1.
CMR: \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2>33\)
Cho a,b,c dương và abc=1
CMR: \(\frac{a^4}{2\left(b+c\right)^2}+\frac{b^4}{2\left(a+c\right)^2}+\frac{c^4}{2\left(a+b\right)^2}+\frac{1}{c^2\left(a+c\right)\left(a+b\right)}+\frac{1}{b^2\left(a+b\right)\left(b+c\right)}+\frac{1}{a^2\left(a+c\right)\left(a+b\right)}\ge\frac{1}{8}\)
Cho a, b, c là các số dương thỏa mãn: a+b+c=1. CMR:
\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2>33\)
Cho a, b, c là các số dương.
CMR:\(\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)}+\frac{a^2}{c\left(c^2+a^2\right)}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
cho a;b;c là các số thực dương thỏa mãn a+b+c=3.CMR:\(\frac{a}{1+\left(b+c\right)^2}+\frac{b}{1+\left(c+a\right)^2}+\frac{c}{1+\left(a+b\right)^2}\le\frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}\)
bài này mà giải theo SOS là hơi bị tuyệt vời nhé =)))
cho a,b,c là số dương : CMR
\(\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Cho các số dương a, b, c thỏa mãn ab+bc+ca=1.
CMR: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3+\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\frac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\frac{\left(c+a\right)\left(c+b\right)}{c^2}}\)
Cho các số thực dương a;b;c thỏa mãn abc=1
CMR
\(\frac{a}{\left(a+1\right)^2}+\frac{b}{\left(b+1\right)^2}+\frac{c}{\left(c+1\right)^2}-\frac{4}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)<=1/4
Cho các số dương a, b, c thỏa mãn ab+bc+ca=1.
CMR: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3+\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\frac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\frac{\left(c+a\right)\left(c+b\right)}{c^2}}\)
\(VT=\frac{ab+bc+ca}{ab}+\frac{ab+bc+ca}{bc}+\frac{ab+bc+ca}{ca}\)
\(=3+\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\)(1)
Theo BĐT AM-GM: \(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\right]\ge\sqrt{\frac{\left(a+b\right)\left(b+c\right)}{b^2}}\)
Tương tự: \(\frac{1}{2}\left[\frac{a\left(b+c\right)}{bc}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(b+c\right)}{c^2}}\)
\(\frac{1}{2}\left[\frac{c\left(a+b\right)}{ab}+\frac{b\left(c+a\right)}{ca}\right]\ge\sqrt{\frac{\left(a+c\right)\left(a+b\right)}{a^2}}\)
Cộng theo vế 3 BĐT trên rồi thay vào 1 ta sẽ thu được đpcm.
Cho các số dương a,b,c t/m a+b+c=1.CMR
\(^{\left(a+\frac{1}{a}\right)^2}+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{100}{3}\)
Ta có: \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)
\(=\left(a^2+b^2+c^2\right)+\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)+6\)
\(\ge\frac{1}{3}\left(a+b+c\right)^2+\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2+6\)
\(\ge\frac{1}{3}\left(a+b+c\right)^2+\frac{1}{3}\left(\frac{9}{a+b+c}\right)^2+6\)
\(=\frac{100}{3}\left(đpcm\right)\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)