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

Question

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.

Answer 1

Since squaring is monotonic, this is equivalent to $$ a^2 + b^2 + c^2 + d^2 + 2\sqrt{a^2+c^2}\sqrt{b^2+d^2} \ge a^2+b^2+c^2+d^2 + 2(ab + cd) $$ which is in turn equivalent to $$ ab + cd \le \sqrt{a^2+c^2}\sqrt{b^2+d^2}, $$ the Cauchy-Schwarz inequality.

Putting it in vector form is even more elegant. For $\mathbf{a} = (a,c)$ and $\mathbf{b} = (b,d)$, this is $$ |\mathbf{a}| + |\mathbf{b}| \ge |\mathbf{a}+\mathbf{b}|, $$ the triangle inequality.

Answer 2

$$\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\ge \:\sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \\ \left( \sqrt{a^2+c^2}+\sqrt{b^2+d^2} \right)^2 \ge \: \left( \sqrt{\left(a+b\right)^2+\left(c+d\right)^2} \right)^2 \\ a^2+c^2+b^2+d^2+2\left( \sqrt{a^2+c^2}\sqrt{b^2+d^2} \right) \ge \: \left(a+b\right)^2+\left(c+d\right)^2 \\ a^2+c^2+b^2+d^2+2\left( \sqrt{(a^2+c^2)(b^2+d^2)} \right) \ge \: a^2+2ab+b^2+c^2+2cd+d^2 \\ 2\left( \sqrt{(a^2+c^2)(b^2+d^2)} \right) \ge \: 2ab+2cd \: \: \: \text{(subtract} \: a^2+b^2+c^2+d^2 \: \text{from both sides)} \\ \sqrt{(a^2+c^2)(b^2+d^2)} \ge \: ab+cd \: \: \text{(divide both sides by 2)} \\ (a^2+c^2)(b^2+d^2) \ge \: (ab+cd)^2 \\ a^2b^2+a^2d^2+c^2b^2+c^2d^2 \ge \: a^2b^2+2abcd+c^2d^2 \\ a^2d^2+c^2b^2 \ge \: 2abcd \: \: \text{(subtract } \: a^2b^2+c^2d^2 \: \text{from both sides)} \\ a^2d^2+c^2b^2 \ge \: 2abcd$$