Minimizing $P=\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}$ when $ab+bc+ca=2.$

Question

Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=2.$ Find the minimum $$P=\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}.$$

By $a=2,b=1,c=0$ I got a value $\dfrac{2}{3}.$

I've tried to prove $$\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}\ge \frac{2}{3},$$which has full expanding expression is$$3[11abc+2abc(a+b+c)+8+4(a+b+c)+2(a^2c+c^2b+b^2a)]$$$$\ge 2[(abc)^2+9abc+3abc(c+a+b)+2(bc^2+ca^2+ab^2)+4(a+b+c)+20].$$It's enough to prove $$2(abc)^2+16\le 4(a+b+c)+15abc+2(ab^2+bc^2+ca^2).\tag{*}$$ After homogenization, it turns out$$4\sqrt{2}(abc)^2+4\sqrt{2}(ab+bc+ca)^3\le 2(a+b+c)\sqrt{(ab+bc+ca)^5}+15abc\sqrt{(ab+bc+ca)^3}+2(ab^2+bc^2+ca^2)\sqrt{(ab+bc+ca)^3},$$or$$\sqrt{(ab+bc+ca)^3}\left[2(a+b+c)(ab+bc+ca)+15abc+2(ab^2+bc^2+ca^2)\right]\ge 4\sqrt{2}(abc)^2+4\sqrt{2}(ab+bc+ca)^3.$$ Let $a=\min{a,b,c}; b=a+u; c=a+v$ where $u,v\ge 0.$

By squaring both side, it is obvious by BW. See Wolfram result.

I'm looking a nice proof for $(*).$

I think RiverLi's approach which verify the yield $ab^2+bc^2+ca^2$ via $p,q,r$ will help but it is much calculate. See his answer for my similar question.

Also, is there another way to solve the starting inequality without full expanding?

I thought of Cauchy-Schwarz inequality $$\sum_{cyc}\frac{a}{a+ab+2}=\sum_{cyc}\frac{a^2(xa+yb+zc)^2}{a(a+ab+2)(xa+yb+zc)^2}\ge \frac{[x(a^2+b^2+c^2)+2(y+z)]^2}{\sum_{cyc}a(a+ab+2)(xa+yb+zc)^2} \ge \frac{2}{3},$$which reduce the degree in some way but I think it is very ugly and it's probably nothing.

Any hint would be desirable.

Interesting. I saw a proof by d8g3n1v9. The construction $\sum_{cyc} a(3b+c-3)^2 \ge 0$ is a way to deal with the form $a^2b + b^2c + c^2a$ (also $\sum_{cyc} a(3b+c-3)^2 = 0$ at the equality condition of the original inequality). I don't know if it helps here. — River Li, Dec 11 '23 at 14:03
@RiverLi I tried with using C-S and this, but it is little difficult. I didn’t want to expand the demoninator. — d8g3n1v9, Dec 11 '23 at 14:16
@RiverLi Th original form is very elegant but I still can not find a nice proof for it. — Tran Ngoc Khuong Trang, Dec 11 '23 at 14:18
@TATAbox However, many inequalities with nice forms may not admit nice proofs (at least currently). — River Li, Dec 11 '23 at 14:24

score 2 · Accepted Answer · answered Dec 11 '23 at 14:46

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I hope there might be better solution than this. Thanks to TATA box, it suffices to prove the below.

$$2\sum_{cyc}ab^2 + 15abc+4\sum_{cyc}a \geq 2(abc)^2+16$$

Step 1

By the equality condition, we start with the following.

$$\sum_{cyc} a(2b+c-2)^2 \geq 0$$

$$\sum_{cyc}a \sum_{cyc} ab = 2 \sum_{cyc} a$$

Subtracting these two, and by simplifying, we get the following.

$$ 2\sum_{cyc} a b^2 + 6abc + 4 \sum_{cyc} a \geq 16 $$

Step 2

By step 1, it suffices to prove the following.

$$9abc \geq 2(abc)^2$$

This is trivial since $abc \leq (\frac{ab+bc+ca}{3})^{3/2} <1 < \frac{9}{2}$ by AM-GM. Therefore, the problem is solved.

Note that the equality holds if and only if $(a,b,c)=(2,1,0),(1,0,2),(0,2,1)$.

answered Dec 11 '23 at 14:46

d8g3n1v9

453

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It is very nice. Your trick works here. – River Li Dec 11 '23 at 14:56
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@RiverLi Unexpected though. I tried to use C-S, but didn’t work. I think I got lucky. Thank you! – d8g3n1v9 Dec 11 '23 at 14:58
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I think your trick works for some more problems in MSE e.g. – River Li Dec 11 '23 at 15:03
@RiverLi Hope it works! Thanks! – d8g3n1v9 Dec 11 '23 at 15:04
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@d8g3n1v9 I check that the following is true. I'll open a topic soon, hope your trick work (I've created it soon) $$a^2+b^2+c^2+abc\ge \frac{3}{8}(a^3b+b^3c+c^3a)+2, \forall a,b,c\ge 0: ab+bc+ca=2.$$ – Tran Ngoc Khuong Trang Dec 12 '23 at 09:37
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@d8g3n1v9 Is there an interesting application? – Tran Ngoc Khuong Trang Dec 14 '23 at 00:05
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@TATAbox Oh, do you mean this problem? I will try to solve when I have some free time. – d8g3n1v9 Dec 14 '23 at 02:24
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@TATAbox It turns out to prove that $\sum_{cyc}ab(a-2b)^2 \geq 4abc(\sum_{cyc}a -2)$. I am stuck. – d8g3n1v9 Dec 14 '23 at 06:46
@TATAbox Would you mind checking this? https://math.stackexchange.com/q/4827419/1241494 – d8g3n1v9 Dec 14 '23 at 09:29
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@d8g3n1v9 Could you please help me out: https://math.stackexchange.com/questions/4827660/showing-sum-cyc-fracaa2b16-ge-frac316-for-a-b-c-ge-0-and-ab ? – Dec 19 '23 at 03:43
@Anonymous I will try but it might be difficult. – d8g3n1v9 Dec 19 '23 at 09:22

Minimizing $P=\frac{a}{a+ab+2}+\frac{b}{b+bc+2}+\frac{c}{c+ca+2}$ when $ab+bc+ca=2.$

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