Question

Giúp em với

Answer 1

Câu 1:

\(\sqrt{a^2+b^2}\ge\dfrac{a+b}{\sqrt{2}}\\ \Leftrightarrow\sqrt{2a^2+2b^2}\ge a+b\\ \Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\\ \Leftrightarrow a^2-2ab+b^2\ge0\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(luôn.đúng\right)\)

Vậy \(\sqrt{a^2+b^2}\ge\dfrac{a+b}{\sqrt{2}}\), dấu \("="\Leftrightarrow a=b\)

Answer 2

Bạn kham khảo

`\sqrt{a^+b^2}>=(a+b)/(\sqrt{2})`

`<=>\sqrt{2a^2+2b^2}>=a+b`

`<=>2a^2+2b^2>=a^2+2ab+b^2`

`<=>a^2-acb+b^2>=0`

`<=>(a-b)^2>=0`(luôn đúng)

Dấu "=" xảy ra khi 

`a=b`