Lời giải:
Ta thấy:
\(\text{VT}=a+2b+c=(a+b+c)+b=1+b(1)\)
Vế phải:
Áp dụng BĐT AM-GM:
\(4(1-a)(1-c)\leq (1-a+1-c)^2=(2-a-c)^2=(1+a+b+c-a-c)^2=(1+b)^2(2)\)
\(\Rightarrow 4(1-a)(1-b)(1-c)\leq (1-b)(1+b)^2\)
Mà : \((1-b)(1+b)^2-(1+b)=(1+b)[(1-b^2)-1]=-b^2(1+b)\leq 0, \forall b\geq 0\)
Do đó: \((1-b)(1+b)^2\leq 1+b(3)\)
Từ (1);(2);(3) ta có đpcm
Dấu bằng xảy ra khi \(a=c=\frac{1}{2}; b=0\)
Cho a;b;c>0 thỏa \(a+b+c=1\)
cmr: \(a+2b+c\ge4\left(1-a\right)\left(1-b\right)\left(1-c\right)\)
Cho a,b,c\(\ge0\).CMR:\(\left(a^2+b^2+c^2\right)^2\ge4\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
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\(\ge\)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 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 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 \(\ge0,a+b+c=1\)
CMR:
ab+ac+bc\(\ge8\left(a^2+b^2+c^2\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\)
CMR: \(\left(1+\frac{a+b+c}{3}\right)^3\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\ge8\sqrt{abc}\) \(\forall a,b,c\ge0\)
1/CMR
a/\(x^4-2x^3+2x^2-2x+1\ge0\forall x\in R\)
b/cho \(a\ge0,b\ge2,a+b+c=3\). CMR : \(a^2+b^2+c^2\le5\)
c/cho a,b,c >0 . CMR : \(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\ge4\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)
2/ cho \(x,y\ge0,x+y=1\). tìm GTLN,GTNN của A =\(x^2+y^2\)
3/ cho x,y>0 .tìm GTNN của B= \(\frac{\left(x+y\right)^2}{x^2+y^2}+\frac{\left(x+y\right)^2}{xy}\)
