# Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$

Let $$a,b,c>0$$. Prove that $$\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$

A idea is to cancel the denominators, but in this case Muirhead don't work because the inequality is only cyclic, not symmetric.

Another idea would be to apply Cauchy reverse technique: $$(1)\iff\sum \left(a-\frac{a^3}{a^2+b^2}\right)\ge \frac12 \sum (2a-b^2/a)\iff \sum\frac{ab^2}{a^2+b^2}\ge \frac12\sum\frac{2a^2-b^2}{a}$$ $$\iff \sum \frac{(ab)^2}{a^3+ab^2}\ge \frac12\sum \frac{2a^2-b^2}a.$$ Now we can apply Cauchy-Schwarz, and the problem reduces to $$\frac{(\sum ab)^2}{\sum a^3+\sum ab^2}\ge \frac12\sum \frac{2a^2-b^2}{a},$$ and at this point I am stuck. Here the only idea is to cancel the denominators, but as I say above it can't work.

• I put some text on the title for convenience. Commented Feb 3, 2022 at 2:52

## 4 Answers

Using Cauchy-Bunyakovsky-Schwarz, \begin{align*} \frac{ab^2}{a^2 + b^2} + \frac{bc^2}{b^2 + c^2} + \frac{ca^2}{c^2 + a^2} \ge \frac{(b + c + a)^2}{\frac{a^2 + b^2}{a} + \frac{b^2 + c^2}{b} + \frac{c^2 + a^2}{c}} = \frac{(a + b + c)^2}{a + b + c + \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}}. \end{align*} It suffices to prove that $$\frac{(a + b + c)^2}{a + b + c + \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}} \ge a + b + c - \frac12\left(\frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}\right).$$ Denote $$p = a + b + c$$ and $$Q = \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}$$. It suffices to prove that $$\frac{p^2}{p + Q} \ge p - \frac12 Q$$ or $$p^2 \ge (p + Q)(p - Q/2) = p^2 + pQ/2 - Q^2/2$$ or $$Q \ge p$$ i.e., $$\frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c} \ge a + b + c$$ which is true using Cauchy-Bunyakovsky-Schwarz (easy).

• I have a question: how to deduce the inequality from $\sum \frac{b^2}{a}\ge \sum a$? (Because $\frac{(\sum a)^2}{\sum a+\sum \frac{b^2}{a}}\le \frac12\sum a$.) Commented Feb 3, 2022 at 8:34
• @user986772 Denote $p = a + b + c$ and $Q = \frac{ b^2}{a} + \frac{ c^2}{b} + \frac{ a^2}{c}$. The inequality is written as $\frac{p^2}{p + Q} \ge p - Q/2$ or $p^2 \ge (p + Q)(p - Q/2)$ or $Q^2/2 \ge pQ/2$. It suffices to prove that $Q \ge p$. Commented Feb 3, 2022 at 12:24
• Very nice proof, +1! Is this a common technique for solving such inequalities? Commented Feb 3, 2022 at 13:53
• @V.S.e.H. Thanks. I think it is not a common technique since in $p^2 \ge (p + Q)(p - Q/2) = p^2 + pQ/2 - Q^2/2$, $p^2$ is cancelled fortunately. Commented Feb 3, 2022 at 14:23
• Another way: just sum up inequalities of the form $b^2/a\ge 2b-a$ (the latter is equivalent to $(a-b)^2\ge 0$). Commented Feb 3, 2022 at 20:32

A funny solution (which seems correct): note that for $$t>0$$ $$\frac{1}{1+t^2} \leq \frac{1}{2}t^2 + \frac{3}{2}(1-t).$$ To prove this note first that $$p(t) = (t^2 - 3t + 3)(t^2 + 1)$$ has no real roots. Next, $$p''(t) = 12t^2 -18t + 8 > 0$$, implying that $$p$$ is strictly convex. Finally $$p'(t) = 4t^3 - 9t^2 + 8t -3$$, and $$p'(1) = 0$$ implying that $$p$$ has a minimum $$2=p(1)\leq p(t)$$.

Next, substitute $$t\mapsto \frac{b}{a}$$ and multiply both sides by $$a$$, then we get $$\frac{a^3}{a^2+b^2}\leq \frac{1}{2}\frac{b^2}{a} + \frac{3}{2}(a - b).$$ Finally cyclically sum, and done.

• Wow. That's very nice. Commented Feb 5, 2022 at 1:11
• @CalvinLin thanks! Commented Feb 5, 2022 at 1:18
• Yes, it is nice. Commented Feb 5, 2022 at 2:58
• @RiverLi Thanks! Do you know of another simpler way to prove the first inequality? Commented Feb 5, 2022 at 15:32
• @V.S.e.H. How about $$\mathrm{RHS} - \mathrm{LHS} = \frac{(t^2 - t + 1)(t - 1)^2}{2 + 2t^2}?$$ Commented Feb 6, 2022 at 1:08

Let me continue your approach. We will use the following simple observation:

Lemma. For positive $$x$$ and $$y$$ the following inequality holds $$\frac{x^2}{y}\ge 2x-y.$$ Proof. It is equivalent to $$(x-y)^2\ge 0$$.

We need to prove that $$\sum_{cyc}\frac{(ab)^2}{a^3+ab^2}\ge \frac{1}{2}\sum_{cyc}\frac{2a^2-b^2}{a}.$$ Note that $$\frac{(ab)^2}{a^3+ab^2}=\frac{b^2}{a+\frac{b^2}{a}}=\frac{1}{4}\cdot\frac{(2b)^2}{a+\frac{b^2}{a}}\ge\frac{1}{4}\left(4b-a-\frac{b^2}{a}\right).$$ Thus, summing up similar inequalities we obtain $$\sum_{cyc}\frac{(ab)^2}{a^3+ab^2}\ge\frac{3}{4}(a+b+c)-\frac{1}{4}\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).$$ It remains to check that $$\frac{3}{4}(a+b+c)-\frac{1}{4}\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)\ge\frac{1}{2}\sum_{cyc}\frac{2a^2-b^2}{a}.$$ Since $$\sum_{cyc}\frac{2a^2-b^2}{a}=2(a+b+c)-\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)$$, the last inequality is equivalent to $$\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\ge a+b+c,$$ which is again a consequence of the lemma: $$\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\ge (2b-a)+(2c-b)+(2a-c)=a+b+c.$$

Comment. Here is a more general form of the lemma: $$\frac{a^x}{b^y}\ge\frac{xa^{x-y}-yb^{x-y}}{x-y}$$ for any $$a,b,x,y>0$$ such that $$x\neq y$$.

• Lemma and its general form is nice. Commented Feb 3, 2022 at 23:58
• Ah, identical to mine. For the comment, it follows from AM-GM since $(x-y) a^x / b^y + y b^{x-y} \geq x a^{x-y}$. Commented Feb 5, 2022 at 1:13

(Just AM-GM is sufficient.)

From OP's work / River Li's solution, the stated inequality is equivalent to $$\sum a - \frac{a^3}{a^2+b^2} \geq \sum a - \frac{1}{2} \sum \frac{b^2}{a}$$, which is:

$$\sum \frac{ab^2 } { a^2 + b^2 } \geq \sum a - \frac{1}{2} \sum \frac{ b^2}{a}$$

Let $$X = \sum \frac{ab^2 } { a^2 + b^2 } , Y = \sum a, Z = \sum \frac{ b^2}{a}$$ for simplicity.
We want to show that $$X \geq Y - \frac{1}{2} Z$$.

Notice that by AM-GM, $$\frac{ 4ab^2 }{ a^2 + b^2 } + \frac{ a^2 + b^2 } { a} \geq 4b, \quad \frac{ b^2}{a} + a \geq 2b.$$

Summing up the cyclic versions gives us

• $$4X + Y + Z \geq 4Y$$
• $$Z + Y \geq 2Y \Rightarrow Z \geq Y$$.
• Hence $$4 X \geq 3Y - Z \geq 4Y - 2Z$$ as desired.

Note: From $$4X \geq 4Y - (Y+Z)$$, the inequality could be strengthened to

$$\sum \frac{a^3}{a^2+b^2} \leq \frac{1}{4} \sum \frac{ a^2+b^2}{a} \leq \sum \frac{1}{2} \frac{ b^2}{a}$$

• It is nice. (+1) Commented Feb 3, 2022 at 23:55
• It should be $b^2/a+a\ge 2b$, right? Commented Feb 4, 2022 at 7:11
• @richrow Indeed, fixed. Commented Feb 4, 2022 at 15:51