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) $

Question

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.

Answer 1

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.

Answer 2

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 $$\sum_{cyc}\frac{(x-y)^2(x^2+y^2-xz-yz)^2}{x^2+y^2}\geq0.$$

Answer 3

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.