Prove $(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-2abc)$ for $a + b + c = 3$

Question

Problem. Prove the following inequality for non-negative real numbers $a,b,c$: If $a+b+c=3$ then: $$(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-2abc)\qquad(1)$$

There are two more variants of the same problem. The next one seems to be the strongest:

$$(a+b^2)(b+c^2)(c+a^2)\le13\qquad(2)$$

...and the third one follows immediately from the second:

$$(a+b^2)(b+c^2)(c+a^2)\le13+abc\qquad(3)$$

I have tried to tackle the first problem in the following way:

$$abc(1+\frac{b^2}{a})(1+\frac{c^2}{b})(1+\frac{a^2}{c})\le13+abc(1-2abc)$$

With the following replacement:

$$u=\frac{b^2}{a}, v=\frac{c^2}{b}, w=\frac{a^2}{c}, uvw=abc$$

...the inequality becomes:

$$(1+u)(1+v)(1+w)\le \frac{13}{uvw}+1-2uvw$$

Also note that $uvw=abc\le\left( \frac{a+b+c}{3}\right)^3=1$

And now what?

Answer 1

We'll prove the following stronger inequality.

Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3$. Prove that: $$(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-6abc).$$ Indeed, we need to prove that $$\frac{13(a+b+c)^6}{729}+\frac{abc(a+b+c)^3}{27}-6a^2b^2c^2\geq$$ $$\geq\frac{(a(a+b+c)+3b^2)(b(a+b+c)+3c^2)(c(a+b+c)+3a^2)}{27}.$$ Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Hence, $$13(a+b+c)^6+27abc(a+b+c)^3-4374a^2b^2c^2-$$ $$-27(a(a+b+c)+3b^2)(b(a+b+c)+3c^2)(c(a+b+c)+3a^2)=$$ $$=2187(u^2-uv+v^2)a^4+54(52u^3+3uv^2+3uv^2+52v^3)a^3+$$ $$+27(29u^4+98u^3v-51u^2v^2+98uv^3+29v^4)a^2+$$ $$+9(17u^5+31u^4v+62u^3v^2+8u^2v^3+58uv^4+17v^5)a+$$ $$+13u^6-3u^5v+33u^4v^2-64u^3v^3-48u^2v^4+78uv^5+13v^6\geq0.$$

Answer 2

Problem. Let $a, b, c \ge 0$ with $a + b + c = 3$. Prove that $$(a+b^2)(b+c^2)(c+a^2)\le13+abc(1-2abc).$$

Proof. WLOG, assume that $c = \min(a, b, c)$.

Using $abc \le 1$, we have $abc(1 - 2abc) \ge -abc$.

Using $c \le 1$, we have $$(b + c^2)(c + a^2) = bc + c^3 + ba^2 + c^2a^2 \le bc + c + ba^2 + ca^2 = (1 + b)c + (3 - a)a^2.$$

It suffices to prove that $$(a + b^2)[(1 + b)c + (3 - a)a^2]\le 13 - abc$$ or $$13 - (a + b^2) (3- a)a^2 - [ab + (a + b^2)(1 + b)]c \ge 0$$ or $$13 - (a + b^2) (3- a)a^2 - [ab + (a + b^2)(1 + b)](3 - a - b) \ge 0. \tag{1}$$

Actually, (1) is true for all $a \ge 0$ and $b\in \mathbb{R}$. Indeed, we have \begin{align*} &13 - (a + b^2) (3- a)a^2 - [ab + (a + b^2)(1 + b)](3 - a - b)\\[6pt] ={}&\frac{1}{36} \left( 3\,ab+6\,{b}^{2}+4\,a-6\,b-16 \right) ^{2}+\frac{1}{36} \left( 6\,{a}^{2}-9\,a+4\,b-6 \right) ^{2}\\[6pt] &\qquad +{\frac {1}{396}}\, \left( 11\,a+18\,b-44 \right) ^{2}+{\frac {7}{99}}\,{b}^{2}+{\frac {1}{64}}\, a{b}^{2} \left( 8\,a-13 \right) ^{2}+{\frac {5}{192}}\,a{b}^{2}. \end{align*} (Note: Alternatively, we can prove (1) by Buffalo Way (BW). )

We are done.