Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3$. Prove that: $$\frac{a}{\sqrt{b^2+c^2}}+\frac{b}{\sqrt{a^2+c^2}}+\frac{c}{\sqrt{a^2+b^2}}\geq\frac{1}{\sqrt{a^2+bc}}+\frac{1}{\sqrt{b^2+ac}}+\frac{1}{\sqrt{c^2+ab}}$$

I tried to use C-S, Holder, SOS, Rearrangement and more, but without some success.

  • $\begingroup$ Is it possible to prove the inequality with 1 instead of a,b,c of each numerator of the 3 terms in the LHS? $\endgroup$ – Raizen Oct 31 '16 at 11:36
  • $\begingroup$ @Raizen It would wrong. Try $c\rightarrow0^+$ and $a=b$ $\endgroup$ – Michael Rozenberg Oct 31 '16 at 13:08
  • $\begingroup$ @MichaelRozenberg $a=b, c\to 0^+$ gives equality, which seems to suggest Schur is at work. $\endgroup$ – Macavity Oct 31 '16 at 16:53
  • $\begingroup$ This is another case of a general set: if positive $a++c=3$, and $1\leq k \leq 5$ then $$\sum_{cyc}\frac{a}{\sqrt{kb^2+c^2}}\geq \sum_{cyc}\frac{1}{\sqrt{ka^2+bc}}$$ Here the $k=1$ end is sharp (for $k<1$ the inequality fails); but the $k=5$ end actually goes to a bit more than $5.15$. $\endgroup$ – Mark Fischler Apr 3 '17 at 23:52
  • $\begingroup$ @Michael Rozenberg Do you know the kulp quartic ? the cartesian equation is : $$y=\frac{a}{\sqrt{a^2+x^2}}$$ or if we make a good substitution on $y$ we have : $$y'=\frac{b}{\sqrt{a^2+x^2}}$$.Maybe it would be helpfull. $\endgroup$ – user448747 Jun 30 '17 at 15:28

The Buffalo Way works although it is an ugly solution.

Squaring both sides of the inequality, we need to prove that $$\sum_{\mathrm{cyc}} \frac{a^2}{b^2+c^2} + \sum_{\mathrm{cyc}} \frac{2ab}{\sqrt{b^2+c^2}\sqrt{c^2+a^2}} \ge \sum_{\mathrm{cyc}} \frac{1}{a^2+bc} + \sum_{\mathrm{cyc}} \frac{2}{\sqrt{a^2+bc}\sqrt{b^2+ca}}.$$ Using AM-GM and GM-HM, we have $\sqrt{b^2+c^2}\sqrt{c^2+a^2} \le \frac{b^2+c^2 + c^2 + a^2}{2}$ and $\sqrt{a^2+bc}\sqrt{b^2+ca} \ge \frac{2(a^2+bc)(b^2+ca)}{a^2+bc + b^2+ca}$. Thus, it suffices to prove that $$\sum_{\mathrm{cyc}} \frac{a^2}{b^2+c^2} + \sum_{\mathrm{cyc}} \frac{4ab}{b^2+c^2 + c^2 + a^2} \ge \sum_{\mathrm{cyc}} \frac{1}{a^2+bc} + \sum_{\mathrm{cyc}} \frac{a^2+bc + b^2+ca}{(a^2+bc)(b^2+ca)}.$$ After homogenization, we need to prove that $$\sum_{\mathrm{cyc}} \frac{a^2}{b^2+c^2} + \sum_{\mathrm{cyc}} \frac{4ab}{b^2+c^2 + c^2 + a^2} \ge \frac{(a+b+c)^2}{9}\Big(\sum_{\mathrm{cyc}} \frac{1}{a^2+bc} + \sum_{\mathrm{cyc}} \frac{a^2+bc + b^2+ca}{(a^2+bc)(b^2+ca)}\Big)$$ or $f(a, b, c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial with degree $18$.

We use the Buffalo Way. WLOG, assume that $c = \min(a,b,c)$.

If $c = 0$, we have $f(a,b,0) = a^3b^3(2a^2-3ab+2b^2)(2a^2+b^2)(a^2+2b^2)(a^2+b^2)(a-b)^2(a+b)^2$. True.

If $c > 0$ and $c \le a \le b$, let $c = 1, \ a = 1+s, \ b= 1+s + t; \ s,t\ge 0$. $f(1+s, 1+s+t, 1)$ is a polynomial in $s, t$ with non-negative coefficients. True.

If $c > 0$ and $c \le b\le a$, let $c=1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$. $f(1+s+t, 1+s, 1)$ is a polynomial in $s, t$ with non-negative coefficients. True.

We are done.


We begin with a first substitution we put :




The initial inequality become :

$$\frac{v-u}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{v+u}{2}}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})^2}}+\frac{\frac{2w-v-u}{2}}{\sqrt{(v-u)^2+(\frac{u+v}{2})^2}}\geq \frac{1}{\sqrt{(v-u)^2+(\frac{2w-v-u}{2})(\frac{v+u}{2})}}+\frac{1}{\sqrt{(\frac{v+u}{2})^2+(\frac{2w-v-u}{2})(v-u)}}\frac{1}{\sqrt{(\frac{2w-v-u}{2})^2+(v-u)(\frac{v+u}{2})}}$$

We get $w+v-u=3$

Now we make a second substitution :




We get this :


The initial inequality become

$$\frac{2\frac{r-p}{2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{r+p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})^2}}+\frac{\frac{2q-r-p}{2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{r+p}{2})^2}}\geq \frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(2\frac{r-p}{2})^2+(\frac{2q-r-p}{2})(\frac{r+p}{2})}}+\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{r+p}{2})^2+(\frac{2q-r-p}{2})(2\frac{r-p}{2})}}\frac{\sqrt{r^2+q^2-p^2}}{\sqrt{(\frac{2q-r-p}{2})^2+(2\frac{r-p}{2})(\frac{r+p}{2})}}$$

We make a last substitution :




With $P\geq 1$ if we put $p\geq q \geq r$

So the initial condition become :


Wich is equivalent to :


The initial inequality become :

$$\frac{2\frac{1-P}{2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{1+P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})^2}}+\frac{\frac{2Q-1-P}{2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{1+P}{2})^2}}\geq \frac{\sqrt{1+Q^2-P^2}}{\sqrt{(2\frac{1-P}{2})^2+(\frac{2Q-1-P}{2})(\frac{1+P}{2})}}+\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{1+P}{2})^2+(\frac{2Q-1-P}{2})(2\frac{1-P}{2})}}\frac{\sqrt{1+Q^2-P^2}}{\sqrt{(\frac{2Q-1-P}{2})^2+(2\frac{1-P}{2})(\frac{1+P}{2})}}$$

So to conclude it sufficient to combine the condition with the inequality to obtain an inequality with one variable .

  • $\begingroup$ I see two variables. How can you make an inequality with one variable? $\endgroup$ – Michael Rozenberg Jul 12 '17 at 10:20
  • $\begingroup$ If you replace the value of $Q$ in the inequality you have an inequality just with $P$ $\endgroup$ – max8128 Jul 12 '17 at 14:53
  • $\begingroup$ Why $w+v-u=3$? I think $w+v-u=a+2b+c$. $\endgroup$ – Michael Rozenberg Jul 12 '17 at 15:06
  • $\begingroup$ Sorry I'm going to correct it . $\endgroup$ – max8128 Jul 12 '17 at 15:16
  • $\begingroup$ Now it would be correct... $\endgroup$ – max8128 Jul 12 '17 at 15:30

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