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In this question, Calvin Lin points out that the OP has made the incorrect assumption that $\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}} \geq \frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}$ because $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2 $ where $a,b,c\ge0$ and $ab+bc+ca+abc=4$. Assuming that this assumption is correct, learning123 provides a solution to the original question. However, this assumption still remains unproven. So my question is how can we prove it?

$2$ approaches come to mind - casework and proving the original question using an approach that doesn't rely on this initial assumption.

I think there are problems with both approaches. With casework, there seem to be many cases to consider which include all the permutations of $a \geq 2$ or $a < 2$, $b \geq 2$ or $b < 2$ and $a \geq b$ or $a < b$. But even if we systematically write down each case, there are many challenges. Firstly, it is not obvious what the range of values for $c$ will be in each case, and even if we have that information, considering whether our assumption is true given these new domains is a challenge in itself. This is probably equivalent to solving many difficult inequality problems just to ultimately solve $1$. The second approach seems more reasonable to me than the first, but it will still require an ingenious strategy that is not immediately obvious to me.

Considering all of this,

How can we prove $\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}} \geq \frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}$ where $a,b,c\ge0$ and $ab+bc+ca+abc=4$?

  • FYI You're asking the same question as Mars. IE The RHS is exactly 1. – Calvin Lin Mar 29 '22 at 18:48
  • @CalvinLin Yes, I am aware. I chose to ask a separate question as all of the above would not fit in a single comment. –  Mar 29 '22 at 18:51
  • For the casework, you just have to deal with WLOG $ a > 2$. Note that we cannot have 2 values both being $> 2$, since that gives $ab > 4$. – Calvin Lin Mar 29 '22 at 18:52
  • @CalvinLin Hmm, didn’t think of that. A solution still isn’t obvious to me but I'll try to play around a bit. –  Mar 29 '22 at 18:56
  • Agreed that there isn't an obvious solution. $\quad$ To clarify, the incorrect assumption is that $ (2-a) / ( a + \sqrt{bc+abc} ) \geq (2-a) / (a+2)$. When we sum up the cyclic versions, it is possible that the differences cancel out. – Calvin Lin Mar 29 '22 at 19:11

1 Answers1

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Alternative proof.

Notice that $$(b+c-a)^2(bc+abc)=[(a+b+c)^2-4a(b+c)]\cdot[4-a(b+c)]\le [2(a+b+c)-2a(b+c)]^2,$$ $$\implies |b+c-a|\cdot\sqrt{bc+abc}\le |2(a+b+c)-2a(b+c)|. $$

Now, $|b+c-a|\ge b+c-a$ and we'll prove $$2(a+b+c)-2a(b+c)\ge 0, $$ or $$a+b+c+bc+abc-4\ge 0.$$

It's easy to get a substitution $a=\dfrac{2x}{y+z};b=\dfrac{2y}{x+z};c=\dfrac{2z}{y+x}$ where $x,y,z\ge 0: xy+yz+zx>0.$

We'll prove $a+b+c\ge ab+bc+ca$ or $$2\sum_{cyc}\frac{x}{y+z}\ge 4\sum_{cyc}\frac{yz}{(y+x)(z+x)},$$ $$\iff \sum_{cyc}x(x+y)(x+z)\ge 2\sum_{cyc}yz(z+y),$$ $$\iff x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(x+z),$$which is Schur or third degree.

Thus, $a+b+c+abc+bc-4\ge ab+bc+ca+ab+bc-4\ge 0.$

And it gives that $$(b+c-a)\sqrt{bc+abc}\le 2(a+b+c)-2a(b+c) \iff 2(b+c)\sqrt{bc+abc}+2a(b+c)\le (a+b+c)(2+\sqrt{bc+abc}),$$ or $$\frac{2+\sqrt{bc+abc}}{a+\sqrt{bc+abc}}\ge \frac{2(b+c)}{a+b+c} \tag{*}.$$ Take cyclic sum on $(*)$ and we're done.

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