1)
Áp dụng BĐT AM - GM, ta có:
\(A=\dfrac{a}{a-1}\times\dfrac{b}{b-1}\)
\(=\dfrac{ab}{ab-a-b+1}\)
\(=\dfrac{ab-2+2}{ab-2}\) (a + b = 3)
\(=1+\dfrac{2}{ab-2}\)
\(\ge1+\dfrac{2}{\dfrac{\left(a+b\right)^2}{4}-2}\)
\(=9\) (a + b = 3)
Dấu "=" xảy ra khi a = b = 1,5
2)
\(B=\left|x-10\right|+\left|x-11\right|+\left|x-12\right|+\left|x-13\right|\)
\(=\left|x-10\right|+\left|11-x\right|+\left|12-x\right|+\left|x-13\right|\)
\(\ge\left|x-10+11-x+12-x+x-13\right|\)
= 0
Dấu "=" xảy ra khi \(\left(x-10\right)\times\left(11-x\right)\times\left(12-x\right)\times\left(x-13\right)\ge0\)
Bảng xét dấu:
\(MinB=0\Leftrightarrow\left[{}\begin{matrix}x\le10\\11\le x\le12\\13\le x\end{matrix}\right.\)
3)
Áp dụng BĐT Cauchy Shwarz, ta có:
\(C^2=\left(1\times\sqrt{8-3x}+1\times\sqrt{3x-2}\right)^2\)
\(\le\left(1+1\right)\left(8-3x+3x-2\right)\)
\(=12\)
\(\Rightarrow C\le2\sqrt{3}\)
Dấu "=" xảy ra khi \(x=\dfrac{5}{3}\)
