A nice 3 variable inhomogeneous asymmetric inequality by TATA box.

Question

This inequality is proposed by TATA box.

Prove that for ${\forall}a,b,c \geq 0$ such that $ab+bc+ca=2$, prove the following inequality.

$$\sum_{cyc}a^2 + abc \geq \frac{3}{8}\sum_{cyc}a^3 b +2$$

Answer 1

Note that the equality holds if and only if $(a,b,c)=(2,1,0),(1,0,2),(0,2,1)$. We prove this by using this condition.

Step 1

Note that the following holds. This is easy to think by the equality condition.

$$\sum_{cyc}ab(a-2b+c)^2 \geq 0$$

$$\bigg( \sum_{cyc}ab \bigg)^2 = 4$$

By homogenizing and clearing, it suffices to prove the following.

$$\sum_{cyc}ab(a-2b+c)^2 \geq abc(3\sum_{cyc}a -8)$$

Therefore, it suffices to prove when $\sum_{cyc}a > \frac{8}{3} $.

Step 2

Our goal is prove that at least one of the following holds.

$$(a-2b+c)^2 > c (3\sum_{cyc}a - 8)$$

$$(b-2c+a)^2 > a (3\sum_{cyc}a - 8)$$

$$(c-2a+b)^2 > b (3\sum_{cyc}a - 8)$$

Or it suffices to prove the following.

$$\sum_{cyc}(2a-b-c)^2 > (\sum_{cyc}a)(3\sum_{cyc}a - 8)$$

By clearing and using $ab+bc+ca=2$, it suffices to prove the below.

$$3\sum_{cyc}a^2 + 8\sum_{cyc}a > 24$$

This is true by $a+b+c > \frac{8}{3}$ and C-S inequality since the following.

$$3\sum_{cyc}a^2 + 8\sum_{cyc}a \geq (\sum_{cyc}a)^2+ \sum_{cyc}a >\frac{256}{9}>24$$

This proves the inequality.