Can I not use QM-AM inequality to solve this?

Question

I was doing my inequality homework and I encountered the following problem:

Show that $\forall a,b,c\in\mathbb{R}^+$, $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge(a+b+c)\sqrt2.$$

I came up a prove using QM-AM inequality, as follows (you can also try to prove with QM-AM inequality for exercise).

But my problem is, can we not use the QM-AM inequality?

Answer 1

Following the hint given in the comments, consider the following diagram:

The inequality's LHS is the sum of lengths of diagonals that have been drawn in the diagram. However, this being a path from corner to opposite corner, it must be at least as long as a straight line, which has length equal to the RHS. This completes the proof.

Answer 2

By CS:

$\left(\sum\limits_{cyc}\sqrt{a^2+b^2}\right)^2=2\sum\limits_{cyc}a^2+2\sum\limits_{cyc}\sqrt{a^2+b^2}\sqrt{c^2+a^2}\geq 2\sum\limits_{cyc}a^2+2\sum\limits_{cyc}(ac+ab)=2(a+b+c)^2$

Answer 3

By using Hölder’s Inequality, $$ \bigl(a^2+b^2\bigr)^{\tfrac{1}{2}}\bigl(1+1\bigr)^{\tfrac{1}{2}}\ge a+b \\ \sqrt{a^2+b^2\mathstrut} \ge \dfrac{a+b}{\sqrt{2}}$$ Similarly for others, and end it yourself

Answer 4

Here is a purely visual proof based on Michal Adamaszek's comment: