If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$

Question

Problem. Let $a,b,c\ge 0: ab+bc+ca=1$. Find Minimum of $P$: $$P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$$ The source of the problem is unknown.

Here is my attempt:

We have: $$P^2=\frac{1}{2a+bc}+\frac{1}{2b+ca}+\frac{1}{2c+ab}+2\sum_{cyc}\frac{1}{\sqrt{(2a+bc)(2b+ca)}}$$ I proved 2 inequalities: $$\frac{1}{2a+bc}+\frac{1}{2b+ca}+\frac{1}{2c+ab}\ge 2 (1)$$ and: $$\frac{\sqrt{2a+bc}+\sqrt{2b+ca}+\sqrt{2c+ab}}{\sqrt{(2a+bc)(2b+ca)(2c+ab)}}\ge \frac{1+2\sqrt{2}}{2}(2)$$

1. Prove (1): $$\frac{1}{2a+bc}+\frac{1}{2b+ca}+\frac{1}{2c+ab}\ge 2$$ $$\iff f(r)=2r^2+r(4p^2-17p+14)+2(2-p)\le 0 (*)$$ which: $$f^{'}(r)=4r+4p^2-17p+14$$

$\bullet:\sqrt{3}\le p\le 2$ which implies: $ r\le \dfrac{p}{9}\implies f^{'}(r)< 4p^2-\dfrac{149}{9}p+14 =(p-2.95)(p-1.18)<0$

Hence, $f(r)$ is decreasing when $\sqrt{3}\le p\le 2$ and using Schur inequality: $r\ge \dfrac{p(4-p^2)}{9} $, it leads to: $$f(r)\le f\left(\frac{p(4-p^2)}{9}\right)=2p^6-32p^5+137p^4+18p^3-580p^2+342p+324$$ $$=(p-2)\left((p-2)(2p^4-28p^3+17p^2+198p+144)+126\right)<0$$

$\bullet: 2\le p\le 4$, we rewrite (*) as: $$ f(r)=-2r^2-r(4p^2-17p+14)+2(p-2)\ge 0$$ Hence: $$ f^{''}(r)=-4<0\implies min f(r)=min \{f(0);f\left(\frac{1}{3p}\right)\}$$ Calculating: $$f(0)=2(p-2)\ge 0$$ $$f\left(\frac{1}{3p}\right)=\frac{-2}{9p^2}-\frac{4p^2-17p+14}{3p}+2(p-2)\ge \frac{11}{8}>0$$

2. Prove (2):

Squaring both side (2), we will prove: $$2(a+b+c)+1+2\sum_{cyc}\sqrt{(2a+bc)(2b+ca)}\ge \left(\frac{1+2\sqrt{2}}{2}\right)^2(2a+bc)(2b+ca)(2c+ab) $$

It seems complicated for me to full proof. Am I on right approach ? If you find any mistake, please tell me why. I also thought of Holder inequality, which gives: $$P^2.\sum_{cyc}(2a+bc)(mb+nc)^3\ge ((m+n)(a+b+c))^3$$ Hope to see more ideas. Thank you!

Answer 1

The inequality $\sum\limits_{cyc}\frac{1}{2a+bc}\geq2$ we can prove by $uvw$ directly.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $f(w^3)\geq0,$ where $$f(w^3)=-2w^6+A(u,v^2)w^3+B(u,v^2)$$ and we see that $f$ is a concave function,

which says it's enough to prove the last inequality in two following cases.

1. $w^3=0$.

Let $c=0$.

Thus, by AM-GM $$\sum_{cyc}\frac{1}{2a+bc}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{ab}=\frac{a+b}{2}+1\geq2;$$ 2. Two variables are equal.

Let $b=a$. Thus, $c=\frac{1-a^2}{2a},$ where $0<a\leq1$ and we need to prove that $$(1-a)(2-5a-a^2+8a^3-2a^4)\geq0,$$ which is true because $$2-5a-a^2+8a^3-2a^4=$$ $$=\frac{2}{25}(5a-3)^2a(1-a)+\frac{1}{25}(90a^3+53a^2-143a+50)\geq$$ $$\geq\frac{1}{25}\left(90a^3+53a^2+3\cdot\frac{50}{3}-143a\right)\geq\frac{a}{25}\left(5\sqrt[5]{90\cdot53\cdot\left(\frac{50}{3}\right)^3}-143\right)>0.$$ Thus, it's enough to prove that: $$\sum_{cyc}\frac{1}{\sqrt{(2a+bc)(2b+ac)}}\geq\frac{1+2\sqrt2}{2},$$ which is hard.

For example, Holder with $\left((2\sqrt2-1)a+b+c\right)^3,$ which saves the case of an equality occurring, leads to a wrong inequality.

I think, the following Holder can help.

Let $ab=z^2$, $ac=y^2$ and $bc=x^2$, where $x$, $y$ and $z$ are positives.

Thus, $x^2+y^2+z^2=1$ and $$\sum_{cyc}\frac{1}{\sqrt{2a+bc}}=\sum_{cyc}\sqrt{\frac{x}{x^3+2yz}}=$$ $$=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{x}{x^3+2yz}}\right)^2\sum\limits_{cyc}x^2(x^3+2yz)\left(y-\sqrt{\frac{yz}{2}}+z\right)^3}{\sum\limits_{cyc}x^2(x^3+2yz)\left(y-\sqrt{\frac{yz}{2}}+z\right)^3}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}\left(2xy-x\sqrt{\frac{yz}{2}}\right)\right)^3}{\sum\limits_{cyc}x^2(x^3+2yz)\left(y-\sqrt{\frac{yz}{2}}+z\right)^3}}\geq\sqrt2+1,$$ where the last inequality seems true and from here we can get a polynomial inequality.

I hope it will help.

Answer 2

There are 2 nice solutions for $$\sum\limits_{cyc}\dfrac{1}{2a+bc} \ge 2.$$

Solution 1:

By AM-GM $$\sum\limits_{cyc}\dfrac{1}{2a+bc} \ge \sum\limits_{cyc}\dfrac{1}{\dfrac{a}{b+c}+a(b+c)+bc} =\sum\limits_{cyc}\dfrac{b+c}{a+b+c}=2.$$ Equality holds iff $a=b=1, c=0$ and per.

Solution 2:

By C-S $$\dfrac{1}{b+c}+\dfrac{a^2}{a} \ge \dfrac{(a+1)^2}{a+b+c}, \ \forall a>0, \;bc \ge 0,$$ or $$\dfrac{1}{b+c}\ge \dfrac{(a+1)^2}{a+b+c}-a=\dfrac{2a+1-ab-ca}{a+b+c}=\dfrac{2a+bc}{a+b+c}, \ \forall a,b,c \ge 0.$$ So $$\dfrac{1}{2a+bc} \ge \dfrac{b+c}{a+b+c}.$$ And the rest is easy.