Question

B1:C/m

a)\(\dfrac{a^2+b^2}{2}\)\(>=ab\)

b)(a+b)\(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)>=4\) (với a>0,b>0)

c)\(a\left(a+2\right)< \left(a+1\right)^2\)

Answer 1

a.

Giả sử: \(\dfrac{a^2+b^2}{2}\ge ab\) ( đúng )

\(\Leftrightarrow a^2+b^2\ge2ab\) 

\(\Leftrightarrow a^2+b^2-2ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( đúng )

Vậy \(\dfrac{a^2+b^2}{2}\ge ab\)

 

 

Answer 2

b.Giả sử: \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\) ( đúng )

\(\Leftrightarrow\left(a+b\right)\left(\dfrac{a+b}{ab}\right)\ge4\)

\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{ab}\ge4\)

\(\Leftrightarrow\left(a+b\right)^2-4ab\ge0\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\) ( luôn đúng )

Vậy \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)

 

Answer 3

Cảm ơn bn ^^