Lời giải:
Lần sau bạn lưu ý, gõ đầy đủ đề, gõ đúng công thức toán để được hỗ trợ tốt hơn.
Bổ sung: $a,b>0$
Xét hiệu:
$a^3+b^3-ab(a+b)=(a+b)(a-b)^2\geq 0$ với mọi $a,b>0$
$\Rightarrow a^3+b^3\geq ab(a+b)$
$\Rightarrow 3(a^3+b^3)\geq 3ab(a+b)$
$\Rightarrow 4(a^3+b^3)\geq a^3+b^3+3ab(a+b)=(a+b)^3$
$\Rightarrow \frac{a^3+b^3}{2}\geq \frac{(a+b)^3}{8}=\left(\frac{a+b}{2}\right)^3$ (đpcm)
Dấu "=" xảy ra khi $a=b$
Cho tam giác ABC. CMR:
1. Với M tùy ý thì aMA2+bMB2+cMC2≥abc
2. 2(a+b+c)(a2+b2+c2) ≥3 (a3+b3+c3+3abc)
Bài 1:Cho 0<=a;b;c<=2.a+b+c=3
CM:3<=a^3+b^3+c^3-3(a-1)(b-1)(c-1)<=9
Bài 2: Cho -1<=a;b;c<=2.a+b+c=0.CM:
a,a^2+b^2+c^2<=6
b,2abc<=a^2+b^2+c^2<=2abc+2
c,a^2+b^2+c^2<=8-abc
Cho a,b,c > 0. CMR:
\(\dfrac{a^3}{a^2+b^2}+\dfrac{b^3}{b^2+c^2}+\dfrac{c^3}{c^2+a^2}\ge\dfrac{a+b+c}{2}\)
Cho a, b, c, d dương. CM:
1) \(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
2) \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b+c}{\sqrt[3]{abc}}\)
3) \(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{d^2}+\frac{d^2}{a^2}\ge\frac{a+b+c+d}{\sqrt[4]{abcd}}\)
4) \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9;a+b+c\le1\)
CMR với mọi a,b,cϵR
\(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\)
a>b>c>0; cmr: a3b2+b3c2+c3a2>a2b3+b2c3+c2a3
1)cho a,b,c >0. \(cmr:\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ca}+\dfrac{1}{c^2+ab}\le\dfrac{a+b+c}{2abc}\)
2) cho a,b,c>0 và a+b+c=1. \(cmr:\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\ge64\)
3) cho a,b,c>0. \(cme:\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\)
4) cho a,b,c>0 .\(cmr:\dfrac{a^3}{b^3}+\dfrac{b^3}{c^3}+\dfrac{c^3}{a^3}\ge\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\)
5)cho a,b,c>0. cmr: \(\dfrac{1}{a\left(a+b\right)}+\dfrac{1}{b\left(b+c\right)}+\dfrac{1}{c\left(c+a\right)}\ge\dfrac{27}{2\left(a+b+c\right)^2}\)
C/m BĐT : \(\frac{5b^3-a^3}{ab+3b^2}+\frac{5c^3-b^3}{bc+3c^2}+\frac{5a^3-c^3}{ca+3a^2}\le a+b+c\)
\(\frac{c+a}{\sqrt{a^2+c^2}}\ge\frac{c+b}{\sqrt{c^2+b^2}};a>b>0,c>\sqrt{ab}\)
cho a,b,c >0 và \(a^2+b^2+c^2=3\) tìm min của biểu thức
\(P=\frac{a^3}{\sqrt{b^2+3}}+\frac{b^3}{\sqrt{c^2+3}}+\frac{c^3}{\sqrt{a^2+3}}\)
