How to prove $\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73} \geq \sqrt{157}$?

$$\quad{\forall x\in \mathbb{R}:\\ \sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73} \geq \sqrt{157}}$$ I want to prove this.I tried to graph it and see whats going on ...https://www.desmos.com/calculator/xgjovvkal6
I also tried to prove it by derivation ,but it become complicated .
Can anybody give me an idea ? I am thankful in advance.

• Where is this inequality from? – Jack Sep 26 '17 at 17:38
• For the first term, do you mean $\sqrt{2\cdot x}$ or $\sqrt{2}\cdot x$? – MPW Sep 26 '17 at 17:43
• @Jack :From an old book, which written by Parviz shariari ,$$\text{ Parviz Shahriari}$$ de.wikipedia.org/wiki/Parviz_Shahriari It is Persian book ,which mean creativity in solving mathematics . – Khosrotash Sep 26 '17 at 17:45
• It's a bit unsatisfying, but considering that all the functions here are well-behaved with minima around 0, the plot is a proof. I'm sure someone will post a neat trick though. – orlp Sep 26 '17 at 17:51
• @MPW $\sqrt{2x}$ would not even be defined for all real $x$ – Hagen von Eitzen Sep 26 '17 at 18:02

There is a nice proof for $x\geq0$.
By Minkowski we obtain: $$\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73}-\sqrt{157}=$$ $$=\sqrt 2\left(\sqrt{x^2} + \sqrt {{x^2} + x + \frac{1}{2}} + \sqrt {{x^2} - 5x + \frac{13}{2}} + \sqrt {{x^2} - 11x + \frac{73}{2}}-\sqrt{\frac{157}{2}}\right)=$$ $$=\sqrt 2\left(\sqrt{x^2} + \sqrt {\left(x+\tfrac{1}{2}\right)^2+\tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{5}{2}\right)^2+ \tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{11}{2}\right)^2+ \tfrac{25}{4}}-\sqrt{\tfrac{157}{2}}\right)\geq$$ $$=\sqrt 2\left(\sqrt{\left(x+x+\tfrac{1}{2}-x+\tfrac{5}{2}-x+\tfrac{11}{2}\right)^2+\left(0+\frac{1}{2}+\frac{1}{2}+\frac{5}{2}\right)^2}-\sqrt{\tfrac{157}{2}}\right)=$$ $$=13-\sqrt{157}>0.$$
• @Khosrotash Tank you! But for $x<0$ my proof is still very ugly. – Michael Rozenberg Sep 26 '17 at 18:16
• :I saw the problem again , there was $$\forall x\geq 0$$ – Khosrotash Sep 26 '17 at 18:24
• I have found mistake in my proof for $x<0$ and I deleted it. – Michael Rozenberg Sep 26 '17 at 20:10