# Prove that $\frac{a}{b+c^2}+\frac{b}{c+a^2}+\frac{c}{a+b^2}\geq \frac{3}{2}$

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!

• Comments are not for extended discussion; this conversation has been moved to chat. May 27, 2021 at 19:56

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.

• 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. Oct 18, 2021 at 14:50
• @Asher2211 The Vasc's inequality it's exactly, that you saw in your link. We have no a generalization. Oct 18, 2021 at 16:35

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.

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 $$a,b,c\geq 0$$

• 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 $a,b,c\geq 0$ Jun 28 at 13:34