Question

cmr nếu a,b \(\in R\) thì \(\left(\frac{a+b}{2}\right)^2\le\frac{a^2+b^2}{2}\)

Answer 1

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

\(\Leftrightarrow2\left(a^2+2ab+b^2\right)\le4a^2+4b^2\)

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

\(\Leftrightarrow2a^2+2b^2-4ab\ge0\)

\(\Leftrightarrow2\left(a^2+b^2-2ab\right)\ge0\Leftrightarrow2\left(a-b\right)^2\ge0\left(lđ\right)\)

\("="\Leftrightarrow a=b\)

Answer 2

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

\(\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+2ab+b^2\)

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

=> đpcm