Các câu hỏi tương tự
cho \(a,b,c>0.\)\(Cmr:\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2\ge\frac{3}{4}\)
Cho \(a,b,c>0.\)\(Cmr:\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+a\right)\left(c^2+a^2\right)}\ge\frac{a+b+c}{4}\)
Cho a,b,c>0 và abc=1. CMR:
\(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
a,Cho \(a,b,c\in\left[0;1\right].CMR:\)
\(\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\ge\frac{3}{3+abc}\)
b,Cho a,b,c>0 thỏa mãn:abc=1
\(CMR:\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\)
Cho các số dương a,b,c CMR ta luôn có đẳng thức sau :
\(\frac{c\left(a^2+b^2\right)^2}{b^3\left(ab+c^2\right)}+\frac{b\left(c^2+a^2\right)^2}{a^3\left(bc+b^2\right)}+\frac{a\left(b^2+c^2\right)^2}{c^3\left(bc+a^2\right)}\ge\frac{2\left(a^2b+b^2c+c^2a\right)}{abc}\)
Cho a,b,c là 3 số thực dương thỏa mãn a3+b3+c3=1
CMR\(\frac{a^2+b^2}{ab\left(a+b\right)^3}+\frac{b^2+c^2}{bc\left(b+c\right)^3}+\frac{c^2+a^2}{ca\left(c+a\right)^3}\ge\frac{9}{4}\)
Cho a;b;c \(\ge0\).CMR \(\frac{a^3}{b\left(b+c\right)}+\frac{b^3}{c\left(c+a\right)}+\frac{c^3}{a\left(a+b\right)}\)\(\ge\frac{1}{2}\left(a+b+c\right)\)
Cho a,b,c0CMR :frac{a^4}{bleft(b+cright)}+frac{b^4}{cleft(c+aright)}+frac{c^4}{aleft(a+bright)}gefrac{1}{2}left(ab+bc+caright)Áp dụng bđt Svac-xo ta có :VTgefrac{left(a^2+b^2+c^2right)^2}{a^2+b^2+c^2+ab+bc+ca}gefrac{left(a^2+b^2+c^2right)^2}{2left(a^2+b^2+c^2right)}frac{a^2+b^2+c^2}{2}gefrac{ab+bc+ca}{2}Dấu - xảy ra abc
Đọc tiếp
Cho \(a,b,c>0\)
CMR :\(\frac{a^4}{b\left(b+c\right)}+\frac{b^4}{c\left(c+a\right)}+\frac{c^4}{a\left(a+b\right)}\ge\frac{1}{2}\left(ab+bc+ca\right)\)
Áp dụng bđt Svac-xo ta có :
\(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+ab+bc+ca}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{a^2+b^2+c^2}{2}\ge\frac{ab+bc+ca}{2}\)
Dấu "-" xảy ra \(< =>a=b=c\)
Cho a,b,c >0 Chứng minh rằng :
\(\frac{c\left(a^2+b^2\right)^2}{b^3\left(ab+c^2\right)}+\frac{b\left(c^2+a^2\right)^2}{a^3\left(ac+b^2\right)}+\frac{a\left(b^2+c^2\right)^2}{c^3\left(bc+a^2\right)}\ge\frac{2\left(a^2b+b^2c+c^2a\right)}{abc}\)