Finding $\small{\min\limits_{a+b+c=3}\sqrt{ab+bc+ca}\left(\frac1{\sqrt{bc+a}}+\frac1{\sqrt{ca+b}}+\frac1{\sqrt{ab+c}}\right).}$

Question

Let $a,b,c\ge 0: a+b+c=3$ and no two them be zero. Find the minimum of P:$$P=\sqrt{ab+bc+ca}\left(\frac1{\sqrt{bc+a}}+\frac1{\sqrt{ca+b}}+\frac{1}{\sqrt{ab+c}}\right).$$

When I set $a=b=c=1$, we will prove$$P\ge \frac{3\sqrt{6}}{2},$$but it is wrong at $(0,1.5,1.5).$

That why I set $a=b=1.5;c=0$, we will prove$$P\ge 1+\sqrt{6},$$but it is wrong at $(0.1,0.1,2.8).$

In general, $P$ is symmetrical expression of $a,b,c$ which means $P$ achieved minimum at two equal variables.

I tried derivative calculating works:

WLOG, assume that $a\ge b\ge c$. Set $t=\frac{b+c}{2}>0; s=\frac{b-c}{2}\ge0$.

Now, we obtain: $b=s+t; c=t-s\ge 0; a=3-s-t-t+s=3-2t$. We will rewrite $P$ as$$P=f(s,t), t\ge s\ge 0.$$ I think we should continue $f'(s,t)\ge 0$ and we just need to check $P$ achieved minimum at $s=0$ or $b=c.$

Am I in right way? I thought of Holder inequality to eliminate radical yields but it seems hard when I was not sure about equality cases.

Hope to see some helps, thank you very much.

Answer 1

$$P=\sqrt{ab+ac+bc}\sum_{cyc}\sqrt{\frac{3}{a^2+ab+ac+3bc}}\rightarrow\sqrt6$$ for $a=b$ and $c\rightarrow+\infty$(we can do it because $\sqrt{ab+ac+bc}\sum\limits_{cyc}\sqrt{\frac{3}{a^2+ab+ac+3bc}}$ is homogeneous already).

We'll prove that $\sqrt6$ it's an infimum.

Indeed, after replacing $a\rightarrow\frac{1}{a}$ and similar we need to prove that: $$\sum_{cyc}\sqrt{\frac{a}{3a^2+ab+ac+bc}}\geq\sqrt{\frac{2}{a+b+c}},$$ which is true because by Holder $$\sum\limits_{cyc}\sqrt{\frac{a}{3a^2+ab+ac+bc}}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{\frac{a}{3a^2+ab+ac+bc}}\right)^2\sum\limits_{cyc}\frac{a^2}{(3a^2+ab+ac+bc)^2}}{\sum\limits_{cyc}\frac{a^2}{(3a^2+ab+ac+bc)^2}}}\geq$$ $$\geq\sqrt{\frac{\left(\sum\limits_{cyc}\frac{a}{3a^2+ab+ac+bc}\right)^3}{\sum\limits_{cyc}\frac{a^2}{(3a^2+ab+ac+bc)^2}}}\geq\sqrt{\frac{2}{a+b+c}},$$ where the last inequality it's just $$\sum_{cyc}a^6b^6(a-b)^4+abc\sum_{sym}(33a^9b^4+221a^8b^5+514a^7b^6)+$$ $$+a^2b^2c^2\sum_{sym}(123a^8b^2+945a^7b^3+2604a^6b^4+1835a^5b^5)+$$ $$+a^3b^3c^3\sum_{sym}(91a^7+1702a^6b+6522a^5b^2+12458a^4b^3)+$$ $$+a^4b^4c^4\sum_{sym}(4429a^4+21794a^3b+14691a^2b^2)+a^5b^5c^5\sum_{cyc}39042a\geq0,$$ which is obvious.