Chứng minh bất đẳng thức
\(1,\frac{a}{b}+\frac{b}{a}\ge2\)
\(2,a^2+b^2+c^2\ge ab+bc+ca\)
\(3,\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(4,\frac{1}{a}+\frac{1}{b}\ge\frac{4}{ab}\left(a,b>0\right)\)
\(5, 3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)
chứng minh bấ đẳng thức sau :
\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Áp dụng BĐT Cô-si:
\(a^2+b^2\ge2ab\)
\(b^2+c^2\ge2bc\)
\(c^2+a^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
Chứng minh các bất đẳng thức sau:
a,\(\left(\frac{a+b}{2}+\frac{c+d}{2}\right)^2\ge\left(a+c\right)\left(b+d\right)\)
b, \(ab+bc+ca\le0\)khi a+b+c=0
Cho a,b,c>0. Chứng minh: \(a^2+b^2+c^2\ge3\left(ab+bc+ca\right)\) và \(\frac{\left(a+b+c\right)^2}{ab+bc+ca}+\frac{ab+bc+ca}{\left(a+b+c\right)^2}\ge\frac{10}{3}\)
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 \(a,b,c>0\)
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 dương thỏa mãn: \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\ge\left(abc\right)^2\)
CMR: \(CyC\frac{\left(ab\right)^2}{\left(a^2+b^2\right)c^3}\ge\frac{\sqrt{3}}{2}\)
Chứng minh bất đẳng thức
a)\(8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)
b)\(\left(a^2+b^2\right)^2\ge ab\left(a+b\right)^2\)