<s>[INST] Proving $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$ The inequality:
$$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2}$$
But can someone help me with a nice elegant solution. This is an olympiad question I was trying to solve, but couldn't manage an elegant solution.
 [/INST] $$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \\ 
\left( \sqrt{a^2+c^2}+\sqrt{b^2+d^2} \right)^2 \ge \: \left( \sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \right)^2 \\ a^2+c^2+b^2+d^2+2\left( \sqrt{a^2+c^2}\sqrt{b^2+d^2} \right) \ge \: \left(a+b\right)^2+\left(c+d\right)^2 \\ 
a^2+c^2+b^2+d^2+2\left( \sqrt{(a^2+c^2)(b^2+d^2)} \right) \ge \: a^2+2ab+b^2+c^2+2cd+d^2 \\ 2\left( \sqrt{(a^2+c^2)(b^2+d^2)} \right) \ge \: 2ab+2cd \: \: \: \text{(subtract} \: a^2+b^2+c^2+d^2 \: \text{from both sides)} \\ \sqrt{(a^2+c^2)(b^2+d^2)} \ge \: ab+cd \: \: \text{(divide both sides by 2)} \\ (a^2+c^2)(b^2+d^2) \ge \: (ab+cd)^2 \\ a^2b^2+a^2d^2+c^2b^2+c^2d^2 \ge \: a^2b^2+2abcd+c^2d^2 \\ a^2d^2+c^2b^2 \ge \: 2abcd \: \: \text{(subtract } \: a^2b^2+c^2d^2 \: \text{from both sides)} \\ a^2d^2+c^2b^2 \ge \: 2abcd$$
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