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For $a,b,c\ge0: ab+bc+ca+abc=4$ then: $$\frac{2-a}{a+\sqrt{bc+abc}}+\frac{2-b}{b+\sqrt{ca+abc}}+\frac{2-c}{c+\sqrt{ab+abc}}\ge1$$

I used the condition and get: $a+\sqrt{bc+abc}=a+\sqrt{4-a(b+c)}\le a+2$ So we need to prove that: $$\frac{2-a}{a+2}+\frac{2-b}{b+2}+\frac{2-c}{c+2}\ge1$$ I tried to full expand but the rest seems complicated for me. Can anyone help me full my idea? Every thinking is welcomed, thanks!

Sickness
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    Unfortunately, your work is not correct. It is true if $ a \leq 2$, which makes $ 2 - a \geq 0 $. Otherwise, we have $ (2-a) / ( a + \sqrt{bc + abc} ) \leq (2-a) / ( a + 2 ) $. EG If $ a = 4, b = 1 , c = 0$. $\quad$ So while the last line is indeed an equality, you still have more work to be done. – Calvin Lin Mar 29 '22 at 14:08
  • Yes, you are right. I think we can check cases: a>2 and a<2 – Sickness Mar 29 '22 at 14:10
  • As mentioned in a comment, substituting $ c = \frac{ 4-ab }{ ab+a+b}$ into the OP's LHS evaluates to exactly 1. It is left to the reader to check the algebra or trust Wolfram Alpha. – Calvin Lin Mar 29 '22 at 18:50
  • But how we dismiss sqrt ? – Sickness Mar 30 '22 at 04:59
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    To clarify, I'm not referring to your original inequality, and just to the last inequality that you wrote. There are no sqrt there. – Calvin Lin Mar 30 '22 at 22:06
  • Yes, I see now. Thanks. The OP seems hard to me, hope to see more ideas – Sickness Mar 31 '22 at 02:27

2 Answers2

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Let $x = a + 2, \; y = b + 2, \; z = c + 2$

Hence, we need to prove that $$\frac{4-x}{x} + \frac{4-y}{y} + \frac{4-z}{z} \geq 1$$ $$\implies \frac{4}{x} - \frac{x}{x} + \frac{4}{y} - \frac{y}{y} + \frac{4}{z} - \frac{z}{z} \geq 1$$ $$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} - 3 \geq 1$$ $$\implies \frac{4}{x} + \frac{4}{y} + \frac{4}{z} \geq 4$$ $$\implies \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 1$$ $$\implies \frac{1}{a+2} + \frac{1}{b+2} + \frac{1}{c+2} \geq 1$$ $$\implies \frac{4(a + b + c) + ab + bc + ca + 12}{4(a + b + c) + 2(ab + bc + ca) + abc + 8} \geq 1$$ $$\implies 4(a + b + c) + ab + bc + ca + 12 \geq 4(a + b + c) + 2(ab + bc + ca) + abc + 8$$ $$\implies ab + bc + ca +abc -4 \leq 0$$

We know that $ab + bc + ca + abc = 4 \implies ab + bc + ca +abc - 4 = 4 - 4 = 0$.

  • I was just looking at the same analysis. It seems like there should be a simpler way of getting there, as we get perfect equality, but I can't quite see it. – Blitzer Mar 29 '22 at 09:01
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    @Blitzer 1) The first 5 steps can be simplified to "add 1 to each term". 2) We do have an equality, as can be verified that $1/(a+2)+1/(b+2)+1/( (4-ab)/(ab+a+b) +2) =1$. 3) We still have an inequality due to OP's work, with equality holding when $a(b+c) = 0, a=2 $ (and cyclic). – Calvin Lin Mar 29 '22 at 12:17
  • Now, I see my mistake. 2-a>0? – Sickness Mar 29 '22 at 14:09
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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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