# Show that $\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z)$

Show that for positive reals $$x,y,z$$ the following inequality holds and that the constant cannot be improved

$$\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z)$$

Background: I was digging through some old correspondence and found a letter from a very young me to Professor Love at the University of Melbourne. I had apparently ask via a letter (yes it was back when we wrote letters) how one could prove the above inequality (my version had $$1/\sqrt{3}$$ in it). He kindly wrote back but without a full proof. I just found the correspondence today and thought that this was a good question for this site.

Based on his letter and my old writings you can transform the above inequality as follows. First note that $$\text{g.l.b.}f(x,y,z) = \text{g.l.b.}f(x,z,y) = k \quad (say)$$ where g.l.b is the greatest lower bound and $$f(x,y,z)$$ is the function $$\left(\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}}\right)\bigg/(x+y+z).$$ and so $$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq 2k (x+y+z)$$ Thus we need to prove that for positive reals $$x,y$$ and $$x$$ the following is true and tight: $$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \sqrt{2} (x+y+z)$$ One approach to prove this (used by Prof. Love) was to apply Hölder's inequality but this unfortunately only gives: $$\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2\phantom{y}} \geq \frac{2}{\sqrt{3}} (x+y+z)$$ Geometric View: From a geometric view point this inequality can be viewed as stating that the perimeter of $$\Delta PQR$$ is not less than $$\sqrt{2}$$ times the sum of the the three edge-lengths of the box of sides $$x,y,$$ and $$z$$ and the points $$P,Q$$ and $$R$$ are three corners of the box that are not adjacent to each other.

I suspect that this is a "well known" inequality in the right circles but it is still not known to me. Thought that is was a nice problem for lovers of inequalities.

• When $x=y=z$, the inequality is an equality. No improvement possible. Aug 3, 2021 at 16:54
• @herbsteinberg The $𝑧=𝑦=𝑧$ unfortunately does not prove that $\sqrt{2}$ is the best constant. However combining this single case and the partial result proved by Hölder's inequality shows that $2/\sqrt{3} \leq k \leq \sqrt{2}$. I am assuming that is this what you meant.
– vand
Aug 3, 2021 at 18:15

For $$x=y=z$$ the inequality $$\sum\limits_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}\geq k(x+y+z)$$ gives $$k\leq\frac{1}{\sqrt2}$$.

The Peter Scholze's solution for $$k=\frac{1}{\sqrt2}.$$:

By Rearrangement $$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt{y^2+z^2}}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt{x^2+y^2}}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{x^2+y^2}\right)}\geq\frac{x+y+z}{\sqrt2},$$ where the last inequality it's just $$\sum_{cyc}\frac{(x-y)^4}{x^2+y^2}\geq0.$$ In the making of Rearrangement we used the following reasoning.

The triples $$\left(\frac{x^2y^2}{\sqrt{x^2+y^2}},\frac{x^2z^2}{\sqrt{x^2+z^2}},\frac{y^2z^2}{\sqrt{y^2+z^2}}\right)$$ and $$\left(\frac{1}{\sqrt{x^2+y^2}},\frac{1}{\sqrt{x^2+z^2}},\frac{1}{\sqrt{y^2+z^2}}\right)$$ have the opposite ordering, which gives a possibility to use Rearrangement.

• Very nice! I think that some more explanation on how one goes from the equality to inequality (i.e. changing $\sqrt{y^2+z^2}$ to $\sqrt{x^2+y^2}$ )would be very beneficial to many given that the only condition on $x,y$ and $z$ is that they be positive reals. Thanks for the solution
– vand
Aug 3, 2021 at 18:58
• @vand I added something. See now. Aug 3, 2021 at 19:21

Another way.

By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{2(xy+xz+yz)^2}{\sum\limits_{cyc}\frac{x^2+y^2+y^2+z^2}{2}}}=\sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}}$$ and it's enough to prove that: $$\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}\geq\frac{(x+y+z)^2}{2}$$ or $$\sum_{cyc}\left(\frac{x^4}{x^2+y^2}-\frac{3x^2-y^2}{4}\right)+\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}\geq\frac{(x+y+z)^2}{2}-\frac{x^2+y^2+z^2}{2}$$ or $$\sum_{cyc}\frac{(x^2-y^2)^2}{4(x^2+y^2)}\geq xy+xz+yz-\frac{(xy+xz+yz)^2}{x^2+y^2+z^2}$$ or $$\sum_{cyc}(x-y)^2\left(\frac{(x+y)^2}{x^2+y^2}-\frac{2(xy+xz+yz)}{x^2+y^2+z^2}\right)\geq0$$ or $$\sum_{cyc}\frac{(x-y)^2(x^2+y^2-xz-yz)^2}{x^2+y^2}\geq0.$$

• Could you explain how you use CS and AM-Gm-inequalities here to get the inequality $$\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}}\right)}\geq \sqrt{\sum_{cyc}\frac{x^4}{x^2+y^2}+\frac{2(xy+xz+yz)^2}{\sum\limits_{cyc}\frac{x^2+y^2+y^2+z^2}{2}}}$$ The left summands under the square root not change so what we want to show is $$\sum_{cyc} \frac{2x^2y^2}{\sqrt{(x^2+y^2)(y^2+z^2)}} \geq \sum_{cyc} \frac{2(xy+xz+yz)^2}{\sum_{cyc}\frac{x^2+y^2+y^2+z^2}{2}}$$ and from this point I dont know how you argued. Aug 4, 2021 at 18:42
• For example it seems that you next applied AM-GM $\sum_i^n x_i \geq \sqrt[n]{x_1 \cdot ... \cdot x_n}$ for $n=2$ and $x_1= x^2+y^2, x_2= y^2+z^2$ (and their cyclic perputations) to get $$\frac{1}{\sqrt{(x^2+y^2)(y^2+z^2)}} \geq \frac{1}{\frac{x^2+y^2+y^2+z^2}{2}}$$ But from where you obtain the second sum $\sum_{cyc}$ in the demoninator in the right term and what are your vectors $v,w$ when you apply the CS $\langle v,v \rangle \cdot \langle w,w \rangle \geq \langle v,w \rangle^2$? Aug 4, 2021 at 18:43
• @user7391733 Now, by C-S $\sum\limits_{cyc}\frac{x^2y^2}{\frac{x^2+y^2+y^2+z^2}{2}}\geq\frac{(xy+xz+yz)^2}{\sum\limits_{cyc}\frac{x^2+y^2+y^2+z^2}{2}}=\frac{(xy+xz+yz)^2}{2(x^2+y^2+z^2)}.$ Aug 4, 2021 at 19:04
• I see. So you take as $v$ the vector $(\frac{xy}{\sqrt{\frac{x^2+y^2+y^2+z^2}{2}}}, \frac{zx}{\sqrt{\frac{z^2+x^2+x^2+y^2}{2}}}, \frac{yz}{\sqrt{\frac{y^2+z^2+z^2+x^2}{2}}})$ and $w=(\sqrt{\frac{x^2+y^2+y^2+z^2}{2}}, \sqrt{\frac{z^2+x^2+x^2+y^2}{2}}, \sqrt{\frac{y^2+z^2+z^2+x^2}{2}})$ , right? Thank you! Aug 4, 2021 at 19:31
• @user7391733 Yes, of course! Aug 4, 2021 at 21:03

When $$x=y=z$$, lhs=$$\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{x+y+z}{\sqrt{2}}$$=rhs, which means that $$\sqrt{2}$$ is best constant.