a) ta có \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
<=>\(2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)
<=>\(2a^2+2b^2+2c^2\ge2ab+2ac+2cb\)
cộng 2 vế cho \(a^2+b^2+c^2\)ta được điều phải chứng minh
b) Nhân 2 vế cho 9 ta sẽ được:\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)rồi bạn làm tương tự như câu a) là ra nha!
Cho a,b,c >0 Chứng minh rằng :
\(\frac{c\left(a^2+b^2\right)^2}{b^3\left(ab+c^2\right)}+\frac{b\left(c^2+a^2\right)^2}{a^3\left(ac+b^2\right)}+\frac{a\left(b^2+c^2\right)^2}{c^3\left(bc+a^2\right)}\ge\frac{2\left(a^2b+b^2c+c^2a\right)}{abc}\)
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 nguyên dương. CMR:
\(\frac{\left(b+c-a\right)^2}{\left(b+c\right)^2+a^2}+\frac{\left(c+a-b\right)^2}{\left(c+a\right)^2+b^2}+\frac{\left(a+b-c\right)^2}{\left(a+b\right)^2+c^2}\ge\frac{3}{5}\)
(Câu 8 HOMC 2007)
cho a,b,c thuộc R+ và abc=1. chứng minh rằng
\(\frac{1}{a^2\left(b+c\right)}+\frac{1}{b^2\left(c+a\right)}+\frac{1}{c^2\left(a+b\right)}\ge\frac{3}{2}\)
Chứng minh rằng
a, \(2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\))
b, \(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
c, \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c\)
Cho a,b,c>0. CMR
\(\frac{a^3}{\left(b+c\right)^2}+\frac{b^3}{\left(c+a\right)^2}+\frac{c^3}{\left(a+b\right)^2}\ge\frac{a+b+c}{4}\)
Cho a,b,c > 0 và \(a+b+c\le3\)
CMR:\(\frac{a^3}{a^2\left(b+2c\right)}+\frac{b^3}{b^2\left(c+2a\right)}+\frac{c^3}{c^2\left(a+2b\right)}>1\)
Mạnh hơn BĐT Nesbitt:
Chứng minh:\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}+\frac{\left[\Sigma_{cyc}\left(a+b\right)\left(b+c\right)\right]\left(a-b\right)^2}{2\left(a+c\right)\left(b+c\right)\left[\left(b+a\right)\left(c+a\right)+\left(c+b\right)\left(a+b\right)\right]}\)
Với a, b, c > 0
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 \(< =>a=b=c\)
