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

Question

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.

Answer 1

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).

Answer 2

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.

Answer 3

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 $x\neq y$.

Answer 4

(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.

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

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