\(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\) wrong too

Applying the Cauchy-Schwarz inequality,we have :

\(\left(a^2+c^2\right)\left(b^2+d^2\right)\ge\left(ad+bc\right)^2\)

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

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

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

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

\(=a\left(c+d\right)+b\left(c+d\right)=\left(a+b\right)\left(c+d\right)\)

The equality happens only when ab = cd

@AL No.....It's true ! :v 

Ohhh.....i'm sorry about that @AL :)

Change to : \(\sqrt{\left(a^2+c^2\right)\left(b^2+c^2\right)}\)

\(\sqrt{\left(a^2+d^2\right)\left(b^2+d^2\right)}\) - it's wrong