Let $a,b,c\ge 0: ab+bc+ca>0$ and satisfy $a+b+c+abc=4.$ Prove that$$\frac{1}{\sqrt{a^2+b^2}}+\frac{1}{\sqrt{b^2+c^2}}+\frac{1}{\sqrt{c^2+a^2}}\ge \frac{4+\sqrt{2}}{4}.$$
The equality holds at $(0,2,2)$ so that I can not use AM-GM, Cauchy-Schwarz to kill it as usual symmetrical inequality.
I tried Holder inequality$$\left(\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}\right)^2.\sum_{cyc}(a^2+b^2)[(2\sqrt{2}-1)c+a+b]^3\ge(2\sqrt{2}+1)^3(a+b+c)^3 ,$$which saves equality occur but leads to wrong inequality. Counter example: $a=b=\dfrac{1}{100}$.
I hope Holder works in better yields. I also tried Mixing variables without any success.
Best Answer
Maybe the following will help.
By Holder $$\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}\right)^2\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3}{\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3}}\geq\sqrt{\frac{\left(\sum\limits_{cyc}(\sqrt2a^2+ab)\right)^3}{\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3}}$$ and it's enough to prove that: $$8\left(\sum\limits_{cyc}(\sqrt2a^2+ab)\right)^3\geq(1+2\sqrt2)^2\sum\limits_{cyc}(a^2+b^2)(ab+\sqrt2c^2)^3,$$ which saves the case of an equality occurring and it's true for $b=c=0$ and $a=4$.
I checked also that it's true for $b=a$ and $c=\frac{4-2a}{1+a^2},$ where $0\leq a\leq2$.
The term $(\sqrt2c^2+ab)^3$ we can get by the following way.
Note by Holder $$\left(\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}\right)^2\sum_{cyc}(a^2+b^2)(a^2+b^2+kc^2+mab+nac+nbc)^3\geq$$ $$\geq\left(\sum_{cyc}((2+k)a^2+(m+2n)ab)\right)^3.$$ The equality occurs, when $$\left(\frac{1}{\sqrt{a^2+b^2}},\frac{1}{\sqrt{b^2+c^2}},\frac{1}{\sqrt{c^2+a^2}}\right)||$$ $$||\left((a^2+b^2)(a^2+b^2+kc^2+mab+nac+nbc)^3,(b^2+c^2)(b^2+c^2+ka^2+mbc+nab+nac)^3,(c^2+a^2)(c^2+a^2+kb^2+mac+nab+nbc)^3\right),$$ which gives $$\sqrt{a^2+b^2}(a^2+b^2+kc^2+mab+nac+nbc)=\sqrt{b^2+c^2}(b^2+c^2+ka^2+mbc+nab+nac)=\sqrt{c^2+a^2}(c^2+a^2+kb^2+mac+nab+nbc).$$ For $b=c=0$ and $a=4$ the first equation gives $$4\cdot16=0\cdot16k,$$ which is possible, when coefficients before $a^2$ and $b^2$ in the expression $(a^2+b^2+kc^2+mab+nac+nbc)^3$ are equal to $0$ and for the equality case we obtain: $$\sqrt{a^2+b^2}(c^2+mab+nac+nbc)=\sqrt{b^2+c^2}(a^2+mbc+nab+nac)=\sqrt{c^2+a^2}(b^2+mac+nab+nbc),$$ which for $a=b=2$ and $c=0$ gives: $$2\sqrt2\cdot4m=2(4+4n),$$ which for $n=0$(because with $0$ we obtain something simple) gives $m=\frac{1}{\sqrt2}$ and we got an expression $$(\sqrt2c^2+ab)^3,$$ for which Holder saves the case of an equality occurring.