Question

chứng minh bất đẳng thức \(\left(\frac{a+b}{2}\right)^2>=\frac{a^2+b^2}{2}\)

Answer 1

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

\(BDT\Leftrightarrow\frac{\left(a+b\right)^2}{4}\ge\frac{a^2+b^2}{2}\)

\(\Leftrightarrow4\left(a^2+b^2\right)\ge2\left(a+b\right)^2\)

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

\(\Leftrightarrow2a^2+2b^2\ge a^2+2ab+b^2\)

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

BĐT luôn đúng nên ta có ĐPCM

Answer 2

cho x > 0,y>0

chứng minh bất đẳng thức \(_{\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge4}\)