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My question: Let $a,b,c$ be positive real numbers satisfy $a+b+c=3.$ Prove that $$\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}.$$

I have tried to change the LHS to $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}$$ And using Cauchy–Schwarz inequality for it $$\frac{a^2}{ab+ac^2}+\frac{b^2}{bc+ba^2}+\frac{c^2}{ca+cb^2}\geq \frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}$$ Then because $$ab+ca+ca\leq \frac{(a+b+c)^2}{3}=\frac{3^2}{3}=3,$$ $$\frac{(a+b+c)^2}{ab+bc+ca+ac^2+ba^2+cb^2}\geq \frac{9}{3+ac^2+ba^2+cb^2}$$ Finally, I can't prove $ac^2+ba^2+cb^2\leq 3$ $$ $$ I look forward to your help, thank you!

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    May 27, 2021 at 19:56

3 Answers 3

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By C-S and by the Vasc's inequality we obtain:$$\sum_{cyc}\frac{a}{b+c^2}=\sum_{cyc}\frac{a^3}{a^2b+a^2c^2}\geq\frac{\left(\sum\limits_{cyc}\sqrt{a^3}\right)^2}{\sum\limits_{cyc}(a^2b+a^2b^2)}=$$ $$=\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}a^2b\sum\limits_{cyc}a+3\sum\limits_{cyc}a^2b^2}=\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^3b+a^2bc)}\geq$$ $$\geq\frac{3\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}=\frac{(a+b+c)\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}$$ and it's enough to prove that: $$\frac{(a+b+c)\sum\limits_{cyc}(a^3+2\sqrt{a^3b^3})}{\sum\limits_{cyc}(4a^2b^2+a^2bc)+\frac{1}{3}(a^2+b^2+c^2)^2}\geq\frac{3}{2}$$ or $$\sum_{cyc}\left(a^{4}+2a^{3}b+2a^{3}c+4\sqrt{a^5b^3}+4\sqrt{a^5c^3}+4\sqrt{a^3b^3c^2}-14a^{2}b^{2}-3a^{2}bc\right)\geq0,$$ which is smooth.

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  • $\begingroup$ Can you give links of some site which explains generalized vasc's inequality. I already referred the AOPS which only mentions the case for 3 variables. $\endgroup$
    – Asher2211
    Oct 18, 2021 at 14:50
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    $\begingroup$ @Asher2211 The Vasc's inequality it's exactly, that you saw in your link. We have no a generalization. $\endgroup$ Oct 18, 2021 at 16:35
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Remarks: My proof of (1) is not nice. Hope to see a nice proof of it.

My proof:

Since the inequality is cyclic, WLOG, assume that $c = \min(a, b, c)$.

We split into two cases:

Case 1: $c \ge 1/5$

Using Cauchy-Bunyakovsky-Schwarz inequality, we have $$\mathrm{LHS} \ge \frac{(a + b + c)^2}{a(b + c^2) + b(c + a^2) + c(a + b^2)}.$$

It suffices to prove that $$ab + bc + ca + a^2b + b^2c + c^2 a \le 6 \tag{1}$$ which is true (the proof is given at the end).

Case 2: $c < 1/5$

Using $c^2 \le c < 1/5$, we have $$\mathrm{LHS} \ge \frac{a}{b + c} + \frac{b}{1/5 + a^2} \ge \frac{a}{3 - a} + \frac{3 - a - 1/5}{1/5 + a^2}.$$ It suffices to prove that $$\frac{a}{3 - a} + \frac{3 - a - 1/5}{1/5 + a^2} \ge \frac32$$ or (after clearing the denominators) $$25a^3 - 35a^2 - 53a + 75 \ge 0$$ which is true (actually for all $a\ge 0$). (Hint: Using AM-GM, we have $a^3 + \frac94 a \ge 3a^2$. The rest is easy.)

We are done.


Proof of (1):

We split into two cases:

(1) If $a < 1$ or $b < 1$, using the well-known inequality $a^2b + b^2c + c^2a + abc \le \frac{4}{27}(a + b + c)^3$, it suffices to prove that $$ab + bc + ca + 4 - abc \le 6$$ which is written as $$(1 - a)(b - 1)(b + a - 2) + (ab - a - b)(a + b + c - 3) \ge 0$$ which is true (using $a + b > 2$).

(2) If $a, b \ge 1$, let $a = 1 + u$ and $b = 1 + v$ for $u, v \ge 0$, and the inequality is written as $$-u^3 - 3u^2v + v^3 + u^2 + uv + v^2 \ge 0.$$ From $c = 3 - a - b = 1 - u - v$ and $c \ge 1/5$, we have $u + v \le 4/5$. Using $-u^3 \ge - u^2 \cdot \frac45$, it suffices to prove that $$-u^2\cdot \frac45 - 3u^2v + v^3 + u^2 + uv + v^2 \ge 0$$ or $$(1/5 - 3v)u^2 + v^3 + uv + v^2 \ge 0.$$

If $1/5 -3v \ge 0$, the inequality is true.

If $1/5 -3v < 0$, let $g(u) := (1/5 - 3v)u^2 + v^3 + uv + v^2$. Then $g(u)$ is concave. Note that $0 \le u \le 4/5 - v$. Also, we have $g(0) \ge 0$ and $g(4/5 - v) = \frac{-250v^3 + 625v^2 - 180v + 16}{125} \ge 0$. Thus, $g(u) \ge 0$ on $[0, 4/5 - v]$. The desired result follows.

We are done.

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Partial answer hint :

For $x,y,z\in[0.5,2]$ we have the inequality :

$$\frac{x}{y+z^{2}}-\frac{1}{x+y+z}\frac{x}{y+z}\ge0$$

Summing and using the constraint gives the inequality .

Now we have one variable less than $1/2$ come back further .

Extended comment :

I have almost if finish if $\max({a,b,c})=a\ge 2$ the inequality is obvious . Next we have the following inequalities for $2\geq a\geq 1.5\geq c\ge 0.5\geq b$: $$\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)-\left(\frac{a}{b+c^{2}}+\max\left(\frac{1}{2}b,\frac{1}{2}c\right)\right)\ge 0$$ And $$\frac{a}{b+c^{2}}+\frac{b}{c+a^{2}}+\frac{c}{a+b^{2}}-\left(\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)\right)\ge0$$ And finally for $2\ge a\ge 1.6\geq c\ge 0.5\ge b$ : $$\frac{a}{b+c^2}+\max(0.5b,0.5c)\ge 1.5$$ and $a+b+c=3$ and $a,b,c\geq 0$

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  • $\begingroup$ I finish if $\max({a,b,c})=a\ge 2$ the inequality is obvious . Next we have the following inequalities for $2\geq a\geq 1.5\geq c\ge 0.5\geq b$: $$\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)-\left(\frac{a}{b+c^{2}}+\max\left(\frac{1}{2}b,\frac{1}{2}c\right)\right)\ge 0$$ And $$\frac{a}{b+c^{2}}+\frac{b}{c+a^{2}}+\frac{c}{a+b^{2}}-\left(\frac{a}{b+c^{2}}+\frac{c}{a+b^{2}}+\min\left(\frac{1}{8}b,\frac{1}{8}c\right)\right)\ge0$$ And finally for $2\ge a\ge 1.6\geq c\ge 0.5\ge b$ : $$\frac{a}{b+c^2}+\max(0.5b,0.5c)\ge 1.5$$ and $a+b+c=3$ and $a,b,c\geq 0$ $\endgroup$
    – Erik Satie
    Jun 28 at 13:34

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