Question

Giúp em bài này với ạ :(((

Answer 1

a,Áp dụng BĐT Cosi ta có:

\(a^2+\dfrac{1}{a^2}\ge2\sqrt{a^2.\dfrac{1}{a^2}}=2.1=2\)

dấu "=" xảy ra \(\Leftrightarrow a^2=\dfrac{1}{a^2} \Leftrightarrow a^4=1\Leftrightarrow a=\pm 1\)

 

Answer 2

\(b,\dfrac{a+b}{2}\le\sqrt{\dfrac{a^2+b^2}{2}}\)

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

\(\Leftrightarrow\dfrac{a^2+2ab+b^2}{4}\le\dfrac{a^2+b^2}{2}\)

\(\Leftrightarrow2a^2+4ab+2b^2\le4a^2+4b^2\)

\(\Leftrightarrow-2a^2+4ab-2b^2\le0\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow a=b\)