$a+b+c+abc=4,$ prove $\sum_{cyc}\frac{1}{\sqrt{a^2+b^2}}\ge \frac{4+\sqrt{2}}{4}.$

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.

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.

• It is nice but very hard to think of. Thank you very much. How did you use the term $c^2+Kab$? Jul 16, 2023 at 14:31
• @Ha Diep Xuan It's the Bacteria's method. I am ready to explain. Jul 16, 2023 at 17:01
• It seems interesting. The name "Bacteria" recalls me something.... Could you tell more ? Jul 16, 2023 at 23:59
• Oh, now I see your adding. It is very interesting, thank you very much Jul 18, 2023 at 3:03
• @TATA box I wrote: "Maybe the following will help". I think, $uvw$ will help, but I still don't see a full proof. Jul 27, 2023 at 12:32

Some thoughts.

WLOG, assume that $$c = \min(a, b, c)$$.

Using $$\frac{1}{\sqrt{b^2 + c^2}} \ge \frac{1}{b + c}$$ and $$\frac{1}{\sqrt{c^2 + a^2}} \ge \frac{1}{c + a}$$, it suffices to prove that $$\frac{1}{\sqrt{a^2 + b^2}} + \frac{1}{b + c} + \frac{1}{c + a} \ge 1 + \frac{\sqrt 2}{4}. \tag{1}$$

We can prove that $$\frac{1}{b + c} + \frac{1}{c + a} \ge 1. \tag{2}$$ (The proof of (2) is given at the end.)

If $$a^2 + b^2 \le 8$$, clearly (1) is true.

In the following, assume that $$a^2 + b^2 > 8$$.

We have $$\frac{1}{\sqrt{a^2+b^2}} - \frac{\sqrt 2}{4} = \frac{\frac{1}{a^2+b^2} - \frac18}{\frac{1}{\sqrt{a^2+b^2}} + \frac{\sqrt 2}{4}} \ge \frac{\frac{1}{a^2+b^2} - \frac18}{\frac{1}{8}} = \frac{8}{a^2 + b^2} - 1. \tag{3}$$

Using (3) and $$\frac{1}{b+c} \ge \frac{1}{4 - a}$$ and $$\frac{1}{c + a}\ge \frac{1}{4 - b}$$, it suffices to prove that $$\frac{8}{a^2 + b^2} - 1 + \frac{1}{4 - a} + \frac{1}{4 - b} \ge 1$$ which is true for all $$a, b > 0$$ with $$a + b \le 4$$.

Proof of (2):

If $$ab < 1$$, using $$c \le 1$$ and $$a + b \ge 2$$, we have \begin{align*} \frac{1}{b + c} + \frac{1}{c + a} - 1 &= \frac{-c^2 - (a + b - 2)c - ab + a + b}{(b + c)(c + a)}\\ &\ge \frac{-1 - (a + b - 2) - ab + a + b}{(b + c)(c + a)}\\ &= \frac{1 - ab}{(b + c)(c + a)}\\ &\ge 0. \end{align*}

If $$ab \ge 1$$, using $$a + b + c + abc = 4$$, we have $$a + b + c + c \le 4$$. Then by Cauchy-Bunyakovsky-Schwarz inequality, we have $$\frac{1}{b + c} + \frac{1}{c + a} \ge \frac{4}{(b + c) + (c + a)} \ge 1$$.

We are done.

• (+1) Does Jichen lemma help? It suffices to prove 3 conditions. Jul 31, 2023 at 5:58
• @TATAbox Yes, it is true. Jul 31, 2023 at 6:22