Notice that $\frac{1}{a}-a=\frac{1-a^2}{a}=\frac{b^2+c^2}{a}$. After this substitution, multiply both sides by $\sqrt{abc}$ to get that it suffices to prove $\sum_{cyc}\sqrt{bc(b^2+c^2)}\geq \sqrt{2}(\sum_{cyc}\sqrt{ab}\sqrt{bc})$. Since $\sqrt{b^2+c^2}\geq \sqrt{2bc}$, we get LHS $\geq \sqrt{2}(\sum_{cyc}\sqrt{bc}\sqrt{bc})$ which is by rearrangement $\geq RHS$. Thus $LHS\geq RHS$.

pj294 wrote: Prove, that for any positive real numbers $a, b, c$ who satisfy $a^2+b^2+c^2=1$ the following inequality holds. $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$ Proof of Zhangyunhua: $$\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} =\sqrt{\frac{b^2+c^2}{a}} +\sqrt{\frac{c^2+a^2}{b}} +\sqrt{\frac{a^2+b^2}{c}} \geq\sqrt{\frac{2bc}{a}} +\sqrt{\frac{2ca}{b}} +\sqrt{\frac{2ab}{c}} \geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$$

Prove, that for any positive real numbers $a, b, c$ who satisfy $a^2+b^2+c^2=1$ the following inequality holds. $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{6\sqrt3}\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$

scpajmb wrote: Prove, that for any positive real numbers $a, b, c$ who satisfy $a^2+b^2+c^2=1$ the following inequality holds. $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{6\sqrt3}\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$

scpajmb wrote: Prove, that for any positive real numbers $a, b, c$ who satisfy $a^2+b^2+c^2=1$ the following inequality holds. $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{6\sqrt3}\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$ Solution. For the left inequality, without loss of generality, assume that $a=\min\{a,b,c\}$. Thus we have \begin{align*}&3a^2\le a^2+b^2+c^2=1\Longrightarrow \sqrt3\le\frac{1}{a}. \end{align*}Observing that $0<a\le b$ and $0<a\le c$, we obtain that \begin{align*}\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c}\ge3\sqrt{\frac{1}{a}-a}\ge3\sqrt{\sqrt3-\frac{1}{\sqrt3}}=\sqrt{6\sqrt3}. \end{align*}To show the inequality $\sqrt{6\sqrt3}\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$, we first recall $x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}$. Hence \begin{align*}\sqrt{2}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\le\sqrt{6}\sqrt{a+b+c}\le\sqrt{6}\sqrt{\sqrt{3\left(a^2+b^2+c^2\right)}}=\sqrt{6\sqrt3}. \end{align*}We are done. $\blacksquare$ Note: as @Arqady pointed out, this is a wrong proof. Please allow me to leave this wrong proof here so as to always remind me of the mistake I made.

I think there is a problem with the second step.

arqady wrote: I think there is a problem with the second step. Oops! you are right! thanks.

scpajmb wrote: Prove, that for any positive real numbers $a, b, c$ who satisfy $a^2+b^2+c^2=1$ the following inequality holds. $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{6\sqrt3}\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$

Attachments:

scpajmb wrote: Prove, that for any positive real numbers $a, b, c$ who satisfy $a^2+b^2+c^2=1$ the following inequality holds. $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{6\sqrt3}\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}$ For $0<a<1$, we have $a(1-a^2)\leq \frac{2}{3\sqrt{3}}\Longleftrightarrow \left(a-\frac{1}{\sqrt{3}}\right)^2\left(a+\frac{2}{\sqrt{3}}\right)\geq 0.$ Thus, $\sqrt{\frac{1}{a}-a}=\frac{1-a^2}{\sqrt{a(1-a^2)}}\geq \sqrt{\frac{3\sqrt{3}}{2}}(1-a^2)\ $ etc. , We obtain $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c}\geq \sqrt{\frac{3\sqrt{3}}{2}}\{3-(a^2+b^2+c^2)\}=\sqrt{6\sqrt{3}}.$ On the other hand, since $a+b+c\leq \sqrt{3}\sqrt{a^2+b^2+c^2}=\sqrt{3}$, yielding $\sqrt{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})\leq \sqrt{2}\cdot \sqrt{3}\sqrt{a+b+c}=\sqrt{6\sqrt{3}}$, Therefore, we obtain $\sqrt{\frac{1}{a}-a}+\sqrt{\frac{1}{b}-b}+\sqrt{\frac{1}{c}-c} \geq \sqrt{6\sqrt3}\geq \sqrt{2a}+\sqrt{2b}+\sqrt{2c}.$ Equality is attained $a^2+b^2+c^2=1\ (a,\ b, \ c>0)$ and $a=b=c\Longleftrightarrow a=b=c= \frac{1}{\sqrt{3}}.$