Áp dụng BĐT: \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\left(a^2\right)^2+\left(b^2\right)^2+\left(c^2\right)^2\ge a^2b^2+b^2c^2+c^2a^2\ge ab.bc+bc.ca+ca.ab=abc\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
CMR:\(\forall a,b,c\)ta luôn có \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Cho \(a,b,c>0.\)\(Cmr:\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+a\right)\left(c^2+a^2\right)}\ge\frac{a+b+c}{4}\)
\(Cho:a,b,c\ge0.CMR:3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
Cho a,b,c >0. CMR \(3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
Cho a+b+c=0 CMR
\(a^5.\left(b^2+c^2\right)+b^5.\left(c^2+a^2\right)+c^5.\left(a^2+b^2\right)=\frac{1}{2}.\left(a^3+b^3+c^3\right).\left(a^4+b^4+c^4\right)\)
Cho a,b,c,d thỏa mãn: \(a^2+b^2+\left(a-b\right)^2=c^2+d^2+\left(c-d\right)^2\).
CMR: \(a^4+b^4+\left(a-b\right)^4=c^4+d^4+\left(c-d\right)^4\)
Cho a,b,c > 0, abc=1
CMR:
\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}=\frac{b^3}{\left(1+c\right)\left(1+a\right)}=\frac{c^3}{\left(1+b\right)\left(1+a\right)}\)>=\(\frac{3}{4}\)
Cho\(\hept{\begin{cases}a,b,c>0\\abc>1\end{cases}CMR:}2\left(a^2+b^2+c^2\right)+4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge7\left(a+b+c\right)-3\)
Cho a=b=c CMR: \(\left(a^2+b^2+c^2\right)^2=2\left(a^4+b^4+c^4\right)\)
