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$$ \quad{\forall x\in \mathbb{R}:\\ \sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73} \geq \sqrt{157}}$$ 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.

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  • $\begingroup$ Where is this inequality from? $\endgroup$ – Jack Sep 26 '17 at 17:38
  • $\begingroup$ For the first term, do you mean $\sqrt{2\cdot x}$ or $\sqrt{2}\cdot x$? $\endgroup$ – MPW Sep 26 '17 at 17:43
  • $\begingroup$ @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 . $\endgroup$ – Khosrotash Sep 26 '17 at 17:45
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    $\begingroup$ 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. $\endgroup$ – orlp Sep 26 '17 at 17:51
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    $\begingroup$ @MPW $\sqrt{2x}$ would not even be defined for all real $x$ $\endgroup$ – Hagen von Eitzen Sep 26 '17 at 18:02
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There is a nice proof for $x\geq0$.

By Minkowski we obtain: $$\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73}-\sqrt{157}=$$ $$=\sqrt 2\left(\sqrt{x^2} + \sqrt {{x^2} + x + \frac{1}{2}} + \sqrt {{x^2} - 5x + \frac{13}{2}} + \sqrt {{x^2} - 11x + \frac{73}{2}}-\sqrt{\frac{157}{2}}\right)=$$ $$=\sqrt 2\left(\sqrt{x^2} + \sqrt {\left(x+\tfrac{1}{2}\right)^2+\tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{5}{2}\right)^2+ \tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{11}{2}\right)^2+ \tfrac{25}{4}}-\sqrt{\tfrac{157}{2}}\right)\geq$$ $$=\sqrt 2\left(\sqrt{\left(x+x+\tfrac{1}{2}-x+\tfrac{5}{2}-x+\tfrac{11}{2}\right)^2+\left(0+\frac{1}{2}+\frac{1}{2}+\frac{5}{2}\right)^2}-\sqrt{\tfrac{157}{2}}\right)=$$ $$=13-\sqrt{157}>0.$$

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  • $\begingroup$ wow! very very nice . wish I vote +1000 $\endgroup$ – Khosrotash Sep 26 '17 at 18:11
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    $\begingroup$ @Khosrotash Tank you! But for $x<0$ my proof is still very ugly. $\endgroup$ – Michael Rozenberg Sep 26 '17 at 18:16
  • $\begingroup$ :I saw the problem again , there was $$\forall x\geq 0 $$ $\endgroup$ – Khosrotash Sep 26 '17 at 18:24
  • $\begingroup$ :Yes , please . I am learning ...I always enjoy of your ideas. thank again $\endgroup$ – Khosrotash Sep 26 '17 at 18:31
  • $\begingroup$ I have found mistake in my proof for $x<0$ and I deleted it. $\endgroup$ – Michael Rozenberg Sep 26 '17 at 20:10

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