Các câu hỏi tương tự
Cho các số dương a, b, c thỏa mãn ab+bc+ca=1.
CMR: \(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge3+\sqrt{\frac{\left(a+b\right)\left(a+c\right)}{a^2}}+\sqrt{\frac{\left(b+c\right)\left(b+a\right)}{b^2}}+\sqrt{\frac{\left(c+a\right)\left(c+b\right)}{c^2}}\)
Afrac{a^2+bc}{b+ac}+frac{b^2+ca}{c+ab}+frac{c^2+ab}{a+bc}frac{3left(a^2+bcright)}{left(a+b+cright)b+3ac}+frac{3left(b^2+caright)}{left(a+b+cright)c+3ab}+frac{3left(c^2+abright)}{left(a+b+cright)a+3bc}gefrac{3left(a^2+bcright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(b^2+caright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}+frac{3left(c^2+abright)}{left(a^2+bcright)+left(b^2+caright)+left(c^2+abright)}3
Đọc tiếp
\(A=\frac{a^2+bc}{b+ac}+\frac{b^2+ca}{c+ab}+\frac{c^2+ab}{a+bc}\)
\(=\frac{3\left(a^2+bc\right)}{\left(a+b+c\right)b+3ac}+\frac{3\left(b^2+ca\right)}{\left(a+b+c\right)c+3ab}+\frac{3\left(c^2+ab\right)}{\left(a+b+c\right)a+3bc}\)
\(\ge\frac{3\left(a^2+bc\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(b^2+ca\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}+\frac{3\left(c^2+ab\right)}{\left(a^2+bc\right)+\left(b^2+ca\right)+\left(c^2+ab\right)}=3\)
Cho a,b,c > 0. CMR:
\(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge3\sqrt[3]{\frac{3\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(a+b+c\right)}{\left(ab+bc+ca\right)^2}}\)
a,b,c0andleft(a+bright)left(b+cright)left(a+cright)1.Tìm max của ab+bc+acWe have left(a+bright)left(b+cright)left(a+cright)gefrac{8}{9}left(a+b+cright)left(ab+ab+acright) Leftrightarrow1gefrac{8}{9}left(a+b+cright)left(ab+bc+acright).Leftrightarrowfrac{9}{8}geleft(a+b+cright)left(ab+bc+acright)gesqrt{3left(ab+bc+acright)^3}.Leftrightarrowfrac{81}{64}ge3left(ab+bc+acright)^3Leftrightarrowfrac{27}{64}geleft(ab+bc+acright)^3Leftrightarrowfrac{3}{4}ge ab+bc+acVậy Max là frac{3}{4}.Dấu bằng xảy ra kh...
Đọc tiếp
\(a,b,c>0and\left(a+b\right)\left(b+c\right)\left(a+c\right)=1\).Tìm max của \(ab+bc+ac\)
We have \(\left(a+b\right)\left(b+c\right)\left(a+c\right)\ge\frac{8}{9}\left(a+b+c\right)\left(ab+ab+ac\right)\)
\(\Leftrightarrow1\ge\frac{8}{9}\left(a+b+c\right)\left(ab+bc+ac\right).\)
\(\Leftrightarrow\frac{9}{8}\ge\left(a+b+c\right)\left(ab+bc+ac\right)\ge\sqrt{3\left(ab+bc+ac\right)^3}.\)
\(\Leftrightarrow\frac{81}{64}\ge3\left(ab+bc+ac\right)^3\)
\(\Leftrightarrow\frac{27}{64}\ge\left(ab+bc+ac\right)^3\)
\(\Leftrightarrow\frac{3}{4}\ge ab+bc+ac\)
Vậy Max là \(\frac{3}{4}.\)Dấu bằng xảy ra khi \(a=b=c=\frac{1}{2}.\)
Cho a;b;c>0 thỏa mãn abc=1. CMR:
\(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ac+c+1\right)^2}\ge\frac{1}{a+b+c}\)
Cho a>0 b>0 c>0 thỏa mãn a+b+c=1 tính gt bt
\(P=\sqrt{\frac{\left(a+bc\right)\left(b+ac\right)}{c+ab}}+\sqrt{\frac{\left(c+ab\right)\left(b+ac\right)}{a+bc}}+\sqrt{\frac{\left(c+ab\right)\left(a+bc\right)}{b+ac}}\)
Cho a,b, c >0 và \(\frac{c\left(ab+1\right)^2}{b^2\left(bc+1\right)}=\frac{a\left(bc+1\right)^2}{c^2\left(ca+1\right)}=\frac{b\left(ca+1\right)^2}{a^2\left(ab+1\right)}\) CMR: \(a=b=c\)
cm rằng a,b,c khác nhau thì \(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-c\right)\left(b-a\right)}+\frac{a-b}{\left(c-a\right)\left(c-b\right)}=\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}\)
cho a,b,c >0 thỏa mãn a3bc+b3ac+c3ab=a2+b2+c2
CMR: \(\frac{a}{\left(ab+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{abc}{a+b+c}\)