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$\frac{|x-y|}{\sqrt{x^2+1}\sqrt{y^2+1}}+\frac{|y-z|}{\sqrt{y^2+1}\sqrt{z^2+1}}\geq\frac{|z-x|}{\sqrt{z^2+1}\sqrt{x^2+1}}$

Let $$x,y,z\in\mathbb{R}$$, prove this inequality above.

Note by Gurīdo Cuong
3 weeks, 4 days ago

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Let points $$A,B,C$$ be such that $$A (xy,z), B(yz,x) C(zx,y)$$. Note that from the triangle inequality we have that $\overline{AB} \le \overline{AC}+\overline{BC}$ From the Pythagorean theorem, it follows that $|z-x|\sqrt{y^2+1} \le |x-y|\sqrt{z^2+1}+|y-z|\sqrt{x^2+1}$ Dividing both sides by $\prod_{cyc} \sqrt{x^2+1}$ gives us $\frac{|z-x|}{\sqrt{z^2+1}\sqrt{x^2+1}} \le \frac{|x-y|}{\sqrt{x^2+1}\sqrt{y^2+1}}+\frac{|y-z|}{\sqrt{y^2+1}\sqrt{z^2+1}}$ As desired.

This is my first time seeing such an inequality. Where did you come by it? · 2 weeks, 1 day ago

I came by it while doing some trigonometry exercises · 2 weeks ago