Question

CM: \(\dfrac{a^2+2}{\sqrt{a^2+1}}\ge2\)

vs mọi a

Answer 1

\(\dfrac{a^2+2}{\sqrt{a^2+1}}\ge2\)

\(\Leftrightarrow\dfrac{a^2+2}{\sqrt{a^2+1}}-2\ge0\)

\(\Leftrightarrow\dfrac{a^2+2-2\sqrt{a^2+1}}{\sqrt{a^2+1}}\ge0\)

\(\Leftrightarrow\dfrac{\left(a^2+1\right)-2\sqrt{a^2+1}+1}{\sqrt{a^2+1}}\ge0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{a^2+1}-1\right)^2}{\sqrt{a^2+1}}\ge0\) ( đúng )

\(\Rightarrowđpcm\)

Answer 2

lop 8 BDT (loan len)

lop 9 lai ko biet dung nua. " goi la" biet cach van minh the gioi.

<=>√(a^2+1)+1/√(a^2+1)≥2

co si VT≥2

= khi a^2+1=1;a=0