Question

với a,b,c là các số dương. CMR:

a, (a+b)(\(\dfrac{1}{a} +\dfrac{1}{b}\)) \(\ge\) 4 (1)

b, (a+b+c)(\(\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}\)) \(\ge\) 9 (2)

Answer 1

a: \(\left\{{}\begin{matrix}a+b>=2\sqrt{ab}\\\dfrac{1}{a}+\dfrac{1}{b}>=2\cdot\sqrt{\dfrac{1}{ab}}\end{matrix}\right.\)

\(\Leftrightarrow\left(a+b\right)\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2\sqrt{ab}\cdot2\cdot\sqrt{\dfrac{1}{ab}}=4\)

b: \(a+b+c>=3\sqrt[3]{abc}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}>=3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=3\cdot\dfrac{1}{\sqrt[3]{abc}}\)

Do đó: \(\left(a+b+c\right)\cdot\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)