Cho \(a+b\geq0\). Chứng minh rằng $$\dfrac{a+b}{2}\leq\sqrt{\dfrac{a^2+b^2}{2}}$$
\(\dfrac{a+b}{2}\leq\sqrt{\dfrac{a^2+b^2}{2}}\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2}{4}\leq\dfrac{a^2+b^2}{2}\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2}{4}\leq\dfrac{2a^2+2b^2}{4}\)
\(\Leftrightarrow\dfrac{2a^2+2b^2-a^2-2ab-b^2}{4}\geq0\)
\(\Leftrightarrow\dfrac{a^2+b^2-2ab}{4}\geq0\)
\(\Leftrightarrow\dfrac{(a-b)^2}{4}\geq0\) (đúng)
Vậy bất đẳng thức đã cho cũng đúng