# If $a+b+c=3$ so $\sum\limits_{cyc}\frac{a}{\sqrt{b^2+c^2}}\geq\sum\limits_{cyc}\frac{1}{\sqrt{a^2+bc}}$

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.

• Is it possible to prove the inequality with 1 instead of a,b,c of each numerator of the 3 terms in the LHS? – Raizen Oct 31 '16 at 11:36
• @Raizen It would wrong. Try $c\rightarrow0^+$ and $a=b$ – Michael Rozenberg Oct 31 '16 at 13:08
• @MichaelRozenberg $a=b, c\to 0^+$ gives equality, which seems to suggest Schur is at work. – Macavity Oct 31 '16 at 16:53
• 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 $5.15$. – Mark Fischler Apr 3 '17 at 23:52
• @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. – 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 :

$\frac{b}{2}-a=x-2\epsilon<0=-u$

$\frac{b}{2}+a=y+\epsilon=v$

$c+a=z+\epsilon=w$

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 :

$-u=\frac{-p}{\sqrt{r^2+q^2-p^2}}$

$v=\frac{r}{\sqrt{r^2+q^2-p^2}}$

$w=\frac{q}{\sqrt{r^2+q^2-p^2}}$

We get this :

$-p+r+q=3\sqrt{r^2+q^2-p^2}$

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 :

$r=R$

$-p=-RP$

$q=RQ$

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

So the initial condition become :

$-P+1+Q=3\sqrt{1+Q^2-P^2}$

Wich is equivalent to :

$Q=\frac{3}{8}\sqrt{(9P^2-2P-7)}+\frac{(1-P)}{8}$

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 .

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