Lời giải:
\(abc=(1-a)(1-b)(1-c)\Rightarrow \frac{1-a}{a}.\frac{1-b}{b}.\frac{1-c}{c}=1\)
Đặt \(\left(\frac{1-a}{a};\frac{1-b}{b}; \frac{1-c}{c}\right)=(x,y,z)\Rightarrow (a,b,c)=\left(\frac{1}{x+1}; \frac{1}{y+1}; \frac{1}{z+1}\right)\)
Bài toán trở thành
Cho $x,y,z>0$ thỏa mãn $xyz=1$. CMR:
\(A=\frac{1}{(x+1)^2}+\frac{1}{(y+1)^2}+\frac{1}{(z+1)^2}\geq \frac{3}{4}\)
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Thật vậy:
Áp dụng BĐT Bunhiacopxky:
\((x+1)^2\leq (x+\frac{1}{y})(x+y)\Rightarrow \frac{1}{(x+1)^2}\geq \frac{y}{(xy+1)(x+y)}\)
\((y+1)^2\leq (y+\frac{1}{x})(y+x)\Rightarrow \frac{1}{(y+1)^2}\geq \frac{x}{(xy+1)(x+y)}\)
\(\Rightarrow A\geq \frac{y}{(xy+1)(x+y)}+\frac{x}{(xy+1)(x+y)}+\frac{1}{(z+1)^2}\)
\(A\geq \frac{x+y}{(xy+1)(x+y)}+\frac{1}{(z+1)^2}=\frac{1}{xy+1}+\frac{1}{(z+1)^2}\)
\(A\geq \frac{1}{\frac{1}{z}+1}+\frac{1}{(z+1)^2}=\frac{z^2+z+1}{(z+1)^2}(*)\)
Mà \(\frac{z^2+z+1}{(z+1)^2}-\frac{3}{4}=\frac{(z-1)^2}{4(z+1)^2}\geq 0\Rightarrow \frac{z^2+z+1}{(z+1)^2}\geq \frac{3}{4}(**)\)
Từ \((*); (**)\Rightarrow A\geq \frac{3}{4}\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\Leftrightarrow a=b=c=\frac{1}{2}\)