My solution during the contest: There are $x,y,z\in\mathbb{R}^+$ with $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$. We need to prove \[ \sqrt{\frac{x^2}{y^2+z^2}}+\sqrt{\frac{y^2}{z^2+x^2}}+\sqrt{\frac{z^2}{x^2+y^2}}>2 \]Define $\alpha,\beta,\gamma\in(0,1)$ by $\alpha=\frac{x^2}{x^2+y^2+z^2},\beta=\frac{y^2}{x^2+y^2+z^2},\gamma=\frac{z^2}{x^2+y^2+z^2}$. Note that $\alpha+\beta+\gamma=1$. We need to prove \[ \sqrt{\frac{\alpha}{1-\alpha}}+\sqrt{\frac{\beta}{1-\beta}}+\sqrt{\frac{\gamma}{1-\gamma}}>2 \]Since $\alpha<1$ we have \begin{align*} \alpha(2\alpha-1)^2\geq0\\ \alpha\geq4\alpha^2(1-\alpha)\\ \frac{\alpha}{1-\alpha}\geq4\alpha^2\\ \sqrt{\frac{\alpha}{1-\alpha}}\geq2\alpha \end{align*}Since $\alpha>0$ equality holds if and only if $\alpha=\frac{1}{2}$. We have analogous inequalities for $\beta,\gamma$. Thus \[ \sqrt{\frac{\alpha}{1-\alpha}}+\sqrt{\frac{\beta}{1-\beta}}+\sqrt{\frac{\gamma}{1-\gamma}}\geq2(\alpha+\beta+\gamma)=2 \]Equality can only hold for $\alpha=\beta=\gamma=\frac{1}{2}$ which contradicts $\alpha+\beta+\gamma=1$.

If $(a,b,c)=\left(\frac{1}{t},1,t\right)$, we have $abc=1$ and \[ \lim_{t\to\infty}\frac{1}{a\sqrt{c^2+1}}+\frac{1}{b\sqrt{a^2+1}}+\frac{1}{c\sqrt{b^2+1}}=\lim_{t\to\infty}\frac{2}{\sqrt{1+\frac{1}{t^2}}}+\frac{1}{\sqrt{2}t}=2 \]Thus $2$ can not be replaced by a better constant.

Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$\frac{1}{a\sqrt{8c^2 + 1}} + \frac{1}{b\sqrt{8a^2 + 1}} + \frac{1}{c\sqrt{8b^2+1}} \geq 1$$$$\frac{1}{a\sqrt{7c^2 + 2}} + \frac{1}{b\sqrt{7a^2 +2}} + \frac{1}{c\sqrt{7b^2+2}} \geq 1$$

Romania Team Selection Test 1993: Find the greatest real numbers $A$ such that $$\frac{x}{\sqrt{y^{2}+z^{2}}}+\frac{y}{\sqrt{x^{2}+z^{2}}}+\frac{z}{\sqrt{x^{2}+y^{2}}}> A$$is true for all positve reals numberes $x,y,z$ . For $a,b$ and $c$ positive reals, prove that\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}>2.\]

Since $abc=1$ there exist $x,y,z$ such that $a=\frac{x}{y}, b=\frac{y}{z}$ and $c= \frac{z}{x}$. After making this substitution, the original inequality becomes \[ \sum_{cyc} \sqrt{\frac{x^2}{y^2 +z^2}} > 2. \]Observe that by AM-GM $2\sqrt{\frac{y^2 +z^2}{x^2} } \leqslant \frac{y^2+z^2}{x^2} +1=\frac{x^2+y^2+z^2}{x^2}$. Therefore \[ \sqrt{\frac{x^2}{y^2 +z^2}} \geqslant \frac{2x^2}{x^2+y^2+z^2}\]and hence \[ \sum_{cyc} \sqrt{\frac{x^2}{y^2 +z^2}} \geqslant \sum_{cyc} \frac{2x^2}{x^2+y^2+z^2} = 2. \]It is clear that the equality can not hold, hence the conclusion.

Deligne wrote: Observe that by AM-GM $2\sqrt{\frac{y^2 +z^2}{x^2} } \leqslant \frac{y^2+z^2}{x^2} +1=\frac{x^2+y^2+z^2}{y^2 +z^2}$. How did you add 1 and got that in the RHS?

Directly AM-GM on $\frac{y^2+z^2}{x^2} +1$.

We know that $abc = 1$ so let's do some substitutions along with expanding the whole thing $$\frac{1}{a\sqrt{c^2 + 1}} + \frac{1}{b\sqrt{a^2 + 1}} + \frac{1}{c\sqrt{b^2+1}} > 2$$$\iff \frac{bc\sqrt{(a^2 + 1)(b^2 + 1)} + ca\sqrt{(b^2 + 1)(c^2 + 1)} + ab\sqrt{(c^2 +1)(a^2 + 1)}}{abc\sqrt{(a^2 + 1)(b^2 + 1)(c^2 + 1)}} > 2$ $\iff bc\sqrt{(a^2 + 1)(b^2 + 1)} + ca\sqrt{(b^2 + 1)(c^2 + 1)} + ab\sqrt{(c^2 +1)(a^2 + 1)} > 2abc\sqrt{(a^2 + 1)(b^2 + 1)(c^2 + 1)}$ $\iff \frac{1}{a}\sqrt{(a^2 + 1)(b^2 + 1)} + \frac{1}{b}\sqrt{(b^2 + 1)(c^2 + 1)} + \frac{1}{c}\sqrt{(c^2 +1)(a^2 + 1)} > 2\sqrt{(a^2 + 1)(b^2 + 1)(c^2 + 1)}$

$> 2\sqrt{(a^2 + 1)(b^2 + 1)(c^2 + 1)}$ Now Enough to show that the above result is true Applying AM-GM we get

$\implies \sqrt{\frac{(a^2 + 1)(b^2 + 1)}{a^2}} + \sqrt{\frac{(b^2 + 1)(c^2 + 1)}{b^2}}+ \sqrt{\frac{(c^2 +1)(a^2 + 1)}{c^2}} \ge 3\sqrt[3]{(a^2 + 1)^2 (b^2 + 1)^2 (c^2 + 1)^2}$ Now $3\sqrt[3]{(a^2 + 1)^2 (b^2 + 1)^2 (c^2 + 1)^2} > 2\sqrt{(a^2 + 1)(b^2 + 1)(c^2 + 1)}$ $\iff 3^6 \cdot ((a^2 + 1)^2 (b^2 + 1)^2 (c^2 + 1)^2)^2 > 2^6 \cdot ((a^2 + 1)(b^2 + 1)(c^2 + 1))^3$ (raising to the power 6 both sides) now since $(a^2 + 1)(b^2 + 1)(c^2 + 1) > 1\cdot 1\cdot 1 = 1$ we could divide both sides by $((a^2 + 1)(b^2 + 1)(c^2 + 1))^3$ $\iff 3^6(a^2 + 1)(b^2 + 1)(c^2 + 1) > 2^6$ which is true , hence our required result is true.

$\ge 3\sqrt[3]{(a^2 + 1)^2 (b^2 + 1)^2 (c^2 + 1)^2} > 2\sqrt{(a^2 + 1)(b^2 + 1)(c^2 + 1)}$

$> 2\sqrt{(a^2 + 1)(b^2 + 1)(c^2 + 1)}$ Hence proved $\blacksquare$

quirtt wrote: Deligne wrote: Observe that by AM-GM $2\sqrt{\frac{y^2 +z^2}{x^2} } \leqslant \frac{y^2+z^2}{x^2} +1=\frac{x^2+y^2+z^2}{y^2 +z^2}$. How did you add 1 and got that in the RHS? You are right, I did mess up there. Let me try again. By AM-GM we have \[ 2\sqrt{\frac{y^2 +z^2}{x^2} } \leqslant \frac{y^2+z^2}{x^2} +1 =\frac{x^2+y^2+z^2}{x^2} \]and therefore \[ \sqrt{\frac{x^2}{y^2+z^2}} \geqslant \frac{2x^2}{x^2+y^2+z^2}. \]Hence \[ \sum_{cyc}\sqrt{\frac{x^2}{y^2+z^2}} \geqslant \sum_{cyc} \frac{2x^2}{x^2+y^2+z^2} =2, \]and again the equality cannot hold.

Substitute $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$. Then the inequality is $$\sum_{cyc} \sqrt{\frac{x^2}{y^2+z^2}}>2$$Let $p=x^2$, $q=y^2$, $r=z^2$. Then the inequality is $$\sum_{cyc} \sqrt{\frac{p}{q+r}}>2$$Which is true since $$\sum_{cyc} \sqrt{\frac{p}{q+r}}=\sum_{cyc} \frac{p}{\sqrt{p(q+r)}} \stackrel{AM-GM}{>} \sum_{cyc} \frac{2p}{p+q+r}=2$$The equality when applying AM-GM cannot hold since then $a=b+c$, $b=c+a$, $c=a+b$, which is surely not true.

Substitute same as above to obtain equivalence to: $$\sum_{cyc} \frac{x}{\sqrt{y^2+z^2}}>2$$ Homogenize with $x^2+y^2+z^2=1$, together with substitution $x=sin(u)$, $y=sin(v)$, $z=sin(w)$ to obtain that the inequality is equivalent to: $$tan(u)+tan(v)+tan(w)>2$$Under the condition $sin^2(u)+sin^2(v)+sin^2(w)=1$. Note that $tan(u)\geq{}2sin^2(u)$ because $sin(2u)\leq{}1$, and the result follows. We can't obtain equality in all 3 cases, cause that would imply that $u=v=w=\frac{\pi}{4}$, which would mean that the condition $x^2+y^2+z^2=1$ doesn't hold.

Let $a, b, c$ be positive real numbers such that $a+b+c = 1$. Prove that $$\frac{1}{a\sqrt{c^2 + 1}} + \frac{1}{b\sqrt{a^2 + 1}} + \frac{1}{c\sqrt{b^2+1}} \geq \frac{27}{\sqrt{10}} $$

Tintarn wrote: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$\frac{1}{a\sqrt{c^2 + 1}} + \frac{1}{b\sqrt{a^2 + 1}} + \frac{1}{c\sqrt{b^2+1}} > 2.$$ Let $a=\frac{x^2}{y^2},b=\frac{y^2}{z^2},c=\frac{z^2}{x^2}$. We need to show: $$\sum_{\text{cyc}}\sqrt{\frac a{b+c}}>2.$$This inequality is homogenous in $a,b,c$. WLOG $a+b+c=3$, then consider the function $$f(x)=\sqrt{\frac x{3-x}}.$$We need to show $f(a)+f(b)+f(c)>2$. To that end, claim that $f(x)\ge\frac23x$ for $0<x<3$. Indeed: $$f(x)^2-\frac49x^2=\frac{x\left(2x-3\right)^{2}}{27\left(3-x\right)}>0$$for any such $x$, with equality iff $x=\frac32$. Using this: $$f(a)+f(b)+f(c)\ge\frac23(a+b+c)=2$$with equality iff $a=b=c=\frac32$, which contradicts that $a+b+c=3$. So the inequality is strict, and we're done.

jasperE3 wrote: Tintarn wrote: Let $a, b, c$ be positive real numbers such that $abc = 1$. Prove that $$\frac{1}{a\sqrt{c^2 + 1}} + \frac{1}{b\sqrt{a^2 + 1}} + \frac{1}{c\sqrt{b^2+1}} > 2.$$

The euality case is $x=\frac{3}{2}$.