Question

Chứng minh \(\sqrt{a^{2} +b^{2} }\) ≥ \(\dfrac{a +b}{\sqrt{2}}\) với mọi a; b ≥ 0.

Answer 1

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

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

=>\(2a^2+2b^2>=\left(a+b\right)^2=a^2+2ab+b^2\)

=>\(a^2-2ab+b^2>=0\)

=>\(\left(a-b\right)^2>=0\)(luôn đúng)