Question

Cho a,b,c\(\ge\)0 và a+b+c=1

C/m: \(a+2b+c\ge4\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

Answer 1

Lời giải:

\(a,b,c\geq 0\rightarrow 1-a,1-b,1-c\geq 0\)

Áp dụng BĐT Cauchy ngược dấu:
\((1-a)(1-c)\leq \left(\frac{1-a+1-c}{2}\right)^2=\left(\frac{2-a-c}{2}\right)^2=\left(\frac{1+b}{2}\right)^2\) (do $a+b+c=1$)

Do đó:

\(4(1-a)(1-b)(1-c)\leq 4(1-b)\left(\frac{1+b}{2}\right)^2=(1-b)(1+b)^2=(1+b)(1-b^2)\)

Vì \(b^2\geq 0\Rightarrow 1-b^2\leq 1\Rightarrow (1+b)(1-b^2)\leq 1+b=a+b+c+b=a+2b+c\)

Hay \(4(1-a)(1-b)(1-c)\leq a+2b+c\) (đpcm)

Dấu bằng xảy ra khi \((a,b,c)=(0,5; 0; 0,5)\)