Cho 3 số không âm a, b, c có tổng bằng 1. Chứng minh rằng: a+2b+c≥≥ 4(1-a)(1-b)(1-c)
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Ta có:
\(4\left(1-a\right)\left(1-c\right)\left(1-b\right)\le4\left(1-b\right).\frac{\left(1-a+1-c\right)^2}{4}\)
\(=\left(1-b\right)\left(2-a-c\right)^2=\left(1-b\right)\left(a+2b+c\right)^2\)
\(=\left(1-b\right)\left(a+2b+c\right)\left(a+2b+c\right)\)
\(\le\left(a+2b+c\right).\frac{\left(a+2b+c+1-b\right)^2}{4}\)
\(=\left(a+2b+c\right).\frac{\left(a+b+c+1\right)^2}{4}\)
\(=\left(a+2b+c\right).\frac{4}{4}=a+2b+c\)
Dấu = xảy ra khi:
\(\hept{\begin{cases}1-a=1-c\\a+2b+c=1-b\\a+b+c=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}a=c=\frac{1}{2}\\b=0\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thấy : \(a;b;c\ge0;a+b+c=1\) \(\Rightarrow1-a;1-b;1-c\ge0\)
AD BĐT AM - GM ta được : \(4\left(1-a\right)\left(1-c\right)\le\left(2-a-c\right)^2=\left[2-\left(1-b\right)\right]^2=\left(b+1\right)^2\)
\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(b+1\right)^2=\left(1-b^2\right)\left(b+1\right)\le1.\left(b+1\right)=b+1=b+\left(a+b+c\right)=a+2b+c\)
( đpcm )
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Xét \(VT=a+2b+c=1+b\left(1\right)\)
Áp dụng BĐT AG-GM:
\(4\left(1-a\right)\left(1-c\right)\le\left(1-a+1-c\right)^2=\left(2-a-c\right)^2=\left(1+a+b+c-a-c\right)^2=\left(1+b\right)^2\left(2\right)\)
\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(1+b\right)^2\)
Mà \(\left(1-b\right)\left(1+b\right)^2-\left(1-b\right)=\left(1+b\right)\left(1-b^2-1\right)=-b^2\left(1+b\right)\le0,\forall b\ge0\)
Do đó \(\left(1-b\right)\left(1+b\right)^2\le1+b\left(3\right)\)
Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta có ĐPCM
Dấu "=" \(\Leftrightarrow a=c=\dfrac{1}{2};b=0\)
Áp dụng BĐT Cosi cho 2 số không âm 1-a và 1-c có:
4(1-a)(1-c) =<(1-a+1-c)^2=(1+b)^2
Ta có: 4(1-a)(1-b)(1-c)=<(1+b)^2(1-b)=(1-b^2)(1+b)=<1+b=a+2b+c(đpcm)
Dấu = xảy ra khi b=0;a=c=1/2