Given $a, b, c>0$, prove $\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b)$

Question

Given $a,b,c>0$, prove that $$\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b).$$

My attempt:

I have that $$\frac{a^4}{a+b}+\frac{c^2(a+b)}{4}\geq a^{2}c$$ $$\frac{b^4}{b+c}+\frac{a^2(b+c)}{4}\geq b^{2}a$$ $$\frac{c^4}{c+a}+\frac{b^2(c+a)}{4}\geq c^{2}b$$ Adding them up yields $$\sum \frac{a^4}{a+b}\geq\frac{3}{4}\sum a^2c-\frac{1}{4}\sum a^2b.$$

So it boils down to proving that $a^2b+b^2c+c^2a\leq ab^2+bc^2+ca^2$, which isn't correct.

Could you help me with this problem?

Answer 1

Similar to your inequalities, we have

$$ \frac{ 3a^4}{a+b} + \frac{3}{4}c^2 (a+b) \geq 3a^2c $$ $$ \frac{ a^4}{a+b} + \frac{1}{4} b^2(a+b) \geq a^2b $$ $$ \frac{ a^4}{a+b} + \frac{1}{4} a^2(a+b) \geq a^3 $$

Taking the cyclic summation, we get

$$ \sum \frac{ 5a^4 } { a+b} \geq \sum 2a^2 c + \sum \frac{1}{2}a^3 $$

By rearrangement, we know that $ \sum \frac{1}{2}a^3 \geq \sum\frac{1}{2} a^2 c$, thus,

$$ \sum \frac{ a^4 } { a+b} \geq \sum \frac{1}{2} a^2c, $$

---

The second inequality balances out the $a^2 b$ term which OP had (and is why we have a $3:1$ ratio).
The third inequality balances out the $a^3$ term which the second inequality introduces, and thankfully the other term from the third inequality cancels out with the first.

Answer 2

We need to prove that: $$\sum_{cyc}\left(\frac{a^4}{a+b}-\frac{1}{2}ab^2\right)\geq0$$ or $$\sum_{cyc}\frac{a(2a^3-ab^2-b^3)}{a+b}\geq0$$ or

$$\sum_{cyc}\frac{a(a-b)(2a^2+2ab+b^2)}{a+b}\geq0$$ or

$$\sum_{cyc}\left(\frac{a(a-b)(2a^2+2ab+b^2)}{a+b}-\frac{5}{6}(a^3-b^3)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(7a^2+9ab+5b^2)}{a+b}\geq0$$ and we are done!

Now we see that our inequality is true even for any $a$, $b$ and $c$

such that $a+b>0$, $a+c>0$ and $b+c>0$.

Answer 3

$$\dfrac{a^3}{1+\dfrac{b}{a}}+\dfrac{b^3}{1+\dfrac{c}{b}}+\dfrac{c^3}{1+\dfrac{c}{b}} \ge (a^3+b^3+c^3)\dfrac{1}{1+\dfrac{a^2b+b^2c+c^2a}{a^3+b^3+c^3}}\ge \dfrac{1}{2}(a^2c+b^2a+c^2b)$$

$$\Leftrightarrow 2(a^6+b^5+c^6)+3(a^3b^3+b^3c^3+c^3a^3) \ge \sum_{cyc}(a^5c)+\sum_{cyc}(a^4b^2)+\sum_{cyc}(a^3c^2b)+\sum_{cyc}(a^4bc)+3a^2b^2c^2$$

$$\dfrac{2}{3}(a^6+b^6+c^6)+\dfrac{1}{3}(a^3c^3+b^3a^3+c^3b^3)\ge\sum_{cyc}(a^5c)$$ $$\dfrac{1}{3}(a^6+b^6+c^6)+\dfrac{2}{3}(a^3b^3+b^3c^3+c^3a^3)\ge\sum_{cyc}(a^4b^2)$$ $$\dfrac{1}{3}\sum_{cyc}(a^6+a^3c^3+c^3b^3)\ge\sum_{cyc}(a^3c^2b)$$ $$\dfrac{1}{3}\sum_{cyc}(a^6+a^3c^3+a^3b^3)\ge\sum_{cyc}(a^4bc)$$ $$ \dfrac{1}{3}(a^6+b^6+c^6+2(a^3b^3+b^3c^3+c^3a^3)) \ge 3a^2b^2c^2$$