Inequality with square roots :$\sqrt{a^2+b^2-\sqrt2ab} + \sqrt{b^2+c^2-\sqrt2bc} \geq \sqrt{a^2+c^2} $

Question

Let $a, b, c > 0$. Prove that $\sqrt{a^2+b^2-\sqrt2ab} + \sqrt{b^2+c^2-\sqrt2bc} \geq \sqrt{a^2+c^2} $.

I don't know how to start, please suggest.

Answer 1

Dilemian already gave an excellent geometric solution. Below is an algebraic solution.

Write $$\sqrt{a^2+b^2-\sqrt{2}ab}=\sqrt{\frac{1}{2}a^2+\frac{1}{2}\left(a-\sqrt{2}b\right)^2}$$ and $$\sqrt{b^2+c^2-\sqrt{2}bc}=\sqrt{\frac{1}{2}c^2+\frac{1}{2}\left(\sqrt{2}b-c\right)^2}\,.$$ Using the Triangle Inequality, $$\begin{align} \sqrt{a^2+b^2-\sqrt{2}ab}+\sqrt{b^2+c^2-\sqrt{2}bc}&\geq\sqrt{\frac{1}{2}(a+c)^2+\frac{1}{2}\big((a-\sqrt{2}b)+(\sqrt{2}b-c)\big)^2} \\ &\geq\sqrt{\frac{1}{2}(a+c)^2+\frac{1}{2}(a-c)^2}=\sqrt{a^2+c^2} \,. \end{align}$$ The equality holds if and only if $\displaystyle \frac{1}{a}+\frac{1}{c}=\frac{\sqrt{2}}{b}$.

Answer 2

In the plane $\mathbb{R}^2$, let $p=(a,0),\ q=(\frac{b}{\sqrt{2}},\frac{b}{\sqrt{2}}),\ r=(0,c)$. Then it is easy to see that $$\Vert p \Vert=a,\quad\Vert q \Vert=b,\quad\Vert r \Vert=c$$ and that (by law of cosines or direct calculation) $$\Vert p-q \Vert=\sqrt{a^2+b^2-\sqrt{2}ab},\quad\Vert q-r \Vert=\sqrt{b^2+c^2-\sqrt{2}bc}.$$ Now use the triangle inequality $\Vert p-q \Vert + \Vert q-r \Vert \geq \Vert p-r \Vert$.