the inequality $\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2$

Question

How to show that $$\frac{a^4}{a^3+b^3}+\frac{b^4}{b^3+c^3}+\frac{c^4}{c^3+a^3}\ge \frac{a+b+c}2$$ for $a,b,c>0$?

I tried to prove $$\frac{a^4}{a^3+b^3}\ge \frac {5a}4+\frac{-3b}4$$

but could not continue. Give me ideas, please.

Answer 1

Here is a solution from "Secrets in Inequalities" by Pham Kim Hung.

Answer 2

This one has circular symmetry in $a$, $b$, $c$ so either the largest of the numbers precedes the smallest in circular order, or the smallest precedes the largest.

Two cases:

1. $0< a\le b \le c$

2. $a \ge b \ge c >0$

Bring to a common denominator and you are left to prove that

$$a^7 b^3 + a^6 b^4 - a^4 b^6 + a^3 b^7 - a^6 b^3 c - a^3 b^6 c + a^7 c^3 - a^6 b c^3 - a b^6 c^3 + b^7 c^3 - a^6 c^4 + b^6 c^4 + a^4 c^6 - a^3 b c^6 - a b^3 c^6 - b^4 c^6 + a^3 c^7 + b^3 c^7\ge 0$$

For the first case, make the substitution $a=p$, $b=p+q$, $c=p+q+r$ in the above expression and you get

$$12 p^8 q^2 + 66 p^7 q^3 + 166 p^6 q^4 + 252 p^5 q^5 + 252 p^4 q^6 + 168 p^3 q^7 + 72 p^2 q^8 + 18 p q^9 + 2 q^10 + 12 p^8 q r + 75 p^7 q^2 r + 220 p^6 q^3 r + 408 p^5 q^4 r + 510 p^4 q^5 r + 420 p^3 q^6 r + 216 p^2 q^7 r + 63 p q^8 r + 8 q^9 r + 12 p^8 r^2 + 69 p^7 q r^2 + 183 p^6 q^2 r^2 + 330 p^5 q^3 r^2 + 465 p^4 q^4 r^2 + 468 p^3 q^5 r^2 + 294 p^2 q^6 r^2 + 102 p q^7 r^2 + 15 q^8 r^2 + 30 p^7 r^3 + 129 p^6 q r^3 + 255 p^5 q^2 r^3 + 360 p^4 q^3 r^3 + 400 p^3 q^4 r^3 + 294 p^2 q^5 r^3 + 119 p q^6 r^3 + 20 q^7 r^3 + 40 p^6 r^4 + 141 p^5 q r^4 + 240 p^4 q^2 r^4 + 290 p^3 q^3 r^4 + 240 p^2 q^4 r^4 + 111 p q^5 r^4 + 21 q^6 r^4 + 30 p^5 r^5 + 87 p^4 q r^5 + 129 p^3 q^2 r^5 + 126 p^2 q^3 r^5 + 69 p q^4 r^5 + 15 q^5 r^5 + 12 p^4 r^6 + 27 p^3 q r^6 + 33 p^2 q^2 r^6 + 23 p q^3 r^6 + 6 q^4 r^6 + 2 p^3 r^7 + 3 p^2 q r^7 + 3 p q^2 r^7 + q^3 r^7$$

a polynomial expression in the positive numbers $p$,$q$,$r$ that has all the coefficients $\ge 0$ and hence positive.

In the second case make the substitution $a=p+q+r$, $b=p+q$, $c=p$ and get

$$12 p^8 q^2 + 66 p^7 q^3 + 166 p^6 q^4 + 252 p^5 q^5 + 252 p^4 q^6 + 168 p^3 q^7 + 72 p^2 q^8 + 18 p q^9 + 2 q^10 + 12 p^8 q r + 123 p^7 q^2 r + 444 p^6 q^3 r + 852 p^5 q^4 r + 1002 p^4 q^5 r + 756 p^3 q^6 r + 360 p^2 q^7 r + 99 p q^8 r + 12 q^9 r + 12 p^8 r^2 + 117 p^7 q r^2 + 519 p^6 q^2 r^2 + 1218 p^5 q^3 r^2 + 1695 p^4 q^4 r^2 + 1476 p^3 q^5 r^2 + 798 p^2 q^6 r^2 + 246 p q^7 r^2 + 33 q^8 r^2 + 30 p^7 r^3 + 241 p^6 q r^3 + 807 p^5 q^2 r^3 + 1440 p^4 q^3 r^3 + 1520 p^3 q^4 r^3 + 966 p^2 q^5 r^3 + 343 p q^6 r^3 + 52 q^7 r^3 + 40 p^6 r^4 + 249 p^5 q r^4 + 630 p^4 q^2 r^4 + 850 p^3 q^3 r^4 + 660 p^2 q^4 r^4 + 279 p q^5 r^4 + 49 q^6 r^4 + 30 p^5 r^5 + 135 p^4 q r^5 + 249 p^3 q^2 r^5 + 246 p^2 q^3 r^5 + 129 p q^4 r^5 + 27 q^5 r^5 + 12 p^4 r^6 + 35 p^3 q r^6 + 45 p^2 q^2 r^6 + 31 p q^3 r^6 + 8 q^4 r^6 + 2 p^3 r^7 + 3 p^2 q r^7 + 3 p q^2 r^7 + q^3 r^7$$ again positive.