\(\frac{a^3}{a^2+ab+b^2}=a-\frac{ab\left(a+b\right)}{a^2+ab+b^2}\ge a-\frac{ab\left(a+b\right)}{3ab}=\frac{2a}{3}-\frac{b}{3}\)
Tương tự: \(\frac{b^3}{b^2+bc+c^2}\ge\frac{2b}{3}-\frac{c}{3}\) ; \(\frac{c^3}{c^2+ca+a^2}\ge\frac{2c}{3}-\frac{a}{3}\)
Cộng vế với vế: \(VT\ge\frac{a+b+c}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
cho a,b,c > 0 thỏa mãn ab+bc+ca=1. Cmr:
\(a+b+c+\frac{ab}{b+c}+\frac{bc}{c+a}+\frac{ca}{a+b}\ge\frac{3\sqrt{3}}{2}\)
cho a,b,c> 0 thỏa mãn ab+bc+ca =3. Cmr:
\(\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\ge\frac{3}{4}\)
cho a,b,c > o thỏa mãn ab + bc + ca = 3. Cmr: \(\frac{a^3}{b^2+3}+\frac{b^3}{c^2+3}+\frac{c^3}{a^2+3}\ge\frac{3}{4}\)
cho a,b,c là các số thực dương thỏa mãn ab+bc+ca=3. CMR:
\(\frac{a}{2a^2+bc}+\frac{b}{2b^2+ca}+\frac{c}{2c^2+ab}\ge abc\)
cho a,b,c > 0 .Cmr:
\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)
Cho a, b, c > 0
Cmr: \(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ca+a^2}\)
1, cho a,b,c ≥0 chứng minh các bất đẳng thức sau:
a, (a+b)(b+c)(c+a) ≥ 8abc
b, \(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c,vớia+b+c>0\)
c, \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}vớia,b,c>0\)
cho a,b,c > 0 thỏa mãn a+b+c = 3. Cmr:
\(\frac{a^3}{b^2+c^2}+\frac{b^3}{c^2+a^2}+\frac{c^3}{a^2+b^2}\ge\frac{3}{2}\)
Cho a,b,c>0 và \(a+b+c=\frac{3}{2}\).CMR:
\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\ge\frac{27}{8}\)
