Question

Chứng minh BĐT:

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\)\(\ge\)\(\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)

Answer 1

\(\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2+\left(b+d\right)^2}\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\ge\left(a+c\right)^2+\left(b+d\right)^2\)

\(\Leftrightarrow2\sqrt{a^2+b^2}\sqrt{c^2+d^2}\ge2ac+2bd\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)

BĐT cuối đúng theo BĐT Bunhiacopski 

Dấu "=" khi \(\frac{a}{c}=\frac{b}{d}\)