Question

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

cm BĐT trên

Answer 1

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

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

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

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

   Vì (*) luôn đúng theo bđt bunhia copxki \(\Rightarrow\)đpcm

   dấu ''='' xảy ra khi a/c=b/d

Answer 2

Cái này là Mincopxki rồi bạn. `

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