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

Question

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

Answer 1

There is a nice proof for $x\geq0$.

By Minkowski we obtain: $$\sqrt 2 x + \sqrt {2{x^2} + 2x + 1} + \sqrt {2{x^2} - 10x + 13} + \sqrt {2{x^2} - 22x + 73}-\sqrt{157}=$$ $$=\sqrt 2\left(\sqrt{x^2} + \sqrt {{x^2} + x + \frac{1}{2}} + \sqrt {{x^2} - 5x + \frac{13}{2}} + \sqrt {{x^2} - 11x + \frac{73}{2}}-\sqrt{\frac{157}{2}}\right)=$$ $$=\sqrt 2\left(\sqrt{x^2} + \sqrt {\left(x+\tfrac{1}{2}\right)^2+\tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{5}{2}\right)^2+ \tfrac{1}{4}} + \sqrt {\left(-x+\tfrac{11}{2}\right)^2+ \tfrac{25}{4}}-\sqrt{\tfrac{157}{2}}\right)\geq$$ $$=\sqrt 2\left(\sqrt{\left(x+x+\tfrac{1}{2}-x+\tfrac{5}{2}-x+\tfrac{11}{2}\right)^2+\left(0+\frac{1}{2}+\frac{1}{2}+\frac{5}{2}\right)^2}-\sqrt{\tfrac{157}{2}}\right)=$$ $$=13-\sqrt{157}>0.$$ We'll prove that for $x\leq0$ our inequality is also true.

Indeed, after replacing $x$ at $-x$ we need to prove that: $$-\sqrt 2 x + \sqrt {2{x^2}-2x + 1} + \sqrt {2{x^2}+10x + 13} + \sqrt {2{x^2}+ 22x + 73}>\sqrt{157},$$ where $x\geq0.$

But $$\sqrt{2x^2-2x+1}=\sqrt{2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}}\geq\frac{1}{\sqrt2},$$ $$\sqrt{2x^2+10x+13}-\sqrt2x-\sqrt{11}=\frac{2x^2+10x+13-2x^2-2\sqrt{22}x-11}{\sqrt{2x^2+10x+13}+\sqrt2x+\sqrt{11}}>0$$ and $$\sqrt{2x^2+22x+73}\geq\sqrt{73}.$$ Thus, it's enough to prove that: $$\frac{1}{\sqrt2}+\sqrt{11}+\sqrt{73}>\sqrt{157},$$ which is true.