Lời giải:
Áp dụng BĐT AM-GM:
\(4(1-a)(1-c)\leq (1-a+1-c)^2=(1+b)^2\)
\(\Rightarrow 4(1-a)(1-b)(1-c)\leq (1+b)^2(1-b)(1)\)
Mà:
\(a+2b+c-(1+b)^2(1-b)=1+b-(1+b)(1-b)=(1+b)[1-(1-b^2)]\)
\(=(1+b)b^2>0, \forall b>0\)
\(\Rightarrow a+2b+c> (1+b)^2(1-b)(2)\)
Từ \((1);(2)\Rightarrow a+2b+c> 4(1-a)(1-b)(1-c)\)
cho \(a,b,c\ge0\)và a+b+c=1.
cmr: \(a+2b+c\ge4\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
Cho a,b,c\(\ge\)0 và a+b+c=1
C/m: \(a+2b+c\ge4\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
cho 0<a,b,c<\(\frac{1}{2}\)thỏa mãn a+b+c=1
CMR: \(\frac{1}{a\left(2b+2c-1\right)}+\frac{1}{b\left(2c+2a-1\right)}+\frac{1}{c\left(2a+2b-1\right)}\ge27\)
cho a,b,c>0, CMR:
\(\left(a+b+\dfrac{1}{4}\right)^2+\left(b+c+\dfrac{1}{4}\right)^2+\left(c+a+\dfrac{1}{4}\right)^2\ge4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{a}}\right)\)
Cho a, b, c là độ dài 3 cạnh tam giác. CMR:
1, \(\dfrac{1}{\left(a+b-c\right)^n}+\dfrac{1}{\left(a-b+c\right)^n}+\dfrac{1}{\left(b+c-a\right)^n}\ge\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}\)
2, \(\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}\ge4^n\left[\dfrac{1}{\left(2a+b+c\right)^n}+\dfrac{1}{\left(a+2b+c\right)^n}+\dfrac{1}{\left(a+b+2c\right)^n}\right]\)
Cho a b c là các số thực không âm đôi một khác nhau. CMR :
\(\left(ab+bc+ca\right).\left(\frac{1}{\left(a-b\right)^2}+\frac{1}{\left(b-c\right)^2}\frac{1}{\left(c-a\right)^2}\right)\ge4\)
Cho a,b,c>0. CMR:
\(\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\ge4\left(a+b+c\right)\)
Cho \(\left\{{}\begin{cases}a,b,c>0\\a+2b+3c< 1\end{cases}}\)
CMR: \(\left(1-a\right)\left(1-b\right)^2\left(1-c\right)^3\ge5^6ab^2c^3\)
Cho a,b,c>0 thỏa mãn a+b+c=3 CMR:
\(\dfrac{a^4}{\left(a+2\right)\left(b+2\right)}+\dfrac{b^4}{\left(b+2\right)\left(c+2\right)}+\dfrac{c^4}{\left(c+2\right)\left(a+2\right)}\ge\dfrac{1}{3}\)