For non-negative numbers $a$, $b$ and $c$ such that $ab+bc+ca=1$ find the minimize value of $$P=\frac{1}{\sqrt{a^2+b^2}}+\frac{1}{\sqrt{b^2+c^2}}+\frac{1}{\sqrt{c^2+a^2}}$$
By C-S: $\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)\cdot P\ge\left(1+1+1\right)^2$
And $\sqrt{a^2+b^2}\ge\sqrt{\dfrac{\left(a+b\right)^2}{2}}=\dfrac{a+b}{\sqrt{2}}$
$\Rightarrow\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\ge\dfrac{2\left(a+b+c\right)}{\sqrt{2}}\ge\dfrac{2\cdot\sqrt{3\left(ab+bc+ca\right)}}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}$
$\Rightarrow P\ge\dfrac{\left(1+1+1\right)^2}{\dfrac{2\sqrt{3}}{\sqrt{2}}}=\dfrac{3\sqrt{6}}{2}$
When $a=b=c=\dfrac{1}{\sqrt{3}}$
If $a=0;b=c=1$ it's seem wrong
If $a=b=1$ and $c=0$ then we get a value $2+\frac{1}{\sqrt2}$.
We'll prove that it's a minimal value.
Thus, we need to prove that $$\sum_{cyc}\sqrt{\frac{ab+ac+bc}{a^2+b^2}}\geq2+\frac{1}{\sqrt2}.$$
Indeed, let $c=\min\{a,b,c\}$.
Hence, $$\frac{ab+ac+bc}{a^2+b^2}-\frac{(a+c)(b+c)}{(a+c)^2+(b+c)^2}=\frac{c(a+b+2c)(2ab+ac+bc)}{a^2+b^2)((a+c)^2+(b+c)^2}\geq0,$$ $$\frac{ab+ac+bc}{a^2+c^2}-\frac{b+c}{a+c}=\frac{c(2ab+ac-c^2)}{(a+c)(a^2+c^2)}\geq0$$ and $$\frac{ab+ac+bc}{b^2+c^2}-\frac{a+c}{b+c}=\frac{c(2ab+bc-c^2)}{(b+c)(b^2+c^2)}\geq0.$$ Now, let $\frac{a+c}{b+c}=x^2$ and $\frac{b+c}{a+c}=y^2$, where $x$ and $y$ are positives.
Hence, $xy=1$ and we need to prove that $$x+y+\frac{1}{\sqrt{x^2+y^2}}\geq2+\frac{1}{\sqrt2}$$ or $$x+y-2\sqrt{xy}\geq\frac{1}{\sqrt2}-\frac{1}{\sqrt{x^2+y^2}}$$ or $$(\sqrt{x}-\sqrt{y})^2\geq\frac{(x-y)^2}{\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt2)}$$ or $$\sqrt{2(x^2+y^2)}(\sqrt{x^2+y^2}+\sqrt2)\geq(\sqrt{x}+\sqrt{y})^2.$$ Now, by C-S twice we obtain: $$\sqrt{2(x^2+y^2)}=\sqrt{(1^2+1^2)(x^2+y^2)}\geq x+y=$$ $$=\frac{1}{2}(1^2+1^2)((\sqrt{x})^2+(\sqrt{y})^2)\geq\frac{1}{2}(\sqrt{x}+\sqrt{y})^2.$$ Thus, it's enough to prove that $$\sqrt{x^2+y^2}+\sqrt2\geq2,$$ which is true by AM-GM: $$\sqrt{x^2+y^2}+\sqrt2\geq\sqrt{2xy}+\sqrt2=2\sqrt2>2.$$ Done!