How prove this $ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$

Question

let $a,b,c\ge 0$, show that

$$ b^2c^2+abc(b+c)+a(b^3+c^3)+a^3(b+c)\ge 2a^2(b^2+c^2+bc)$$

my idea use the SOS methods, But I don't work at last. Thank you

Answer 1

Since this expression is symmetric in $b$ and $c$, we can make the substitution $x=b+c$, $y=bc$ to obtain the equivalent inequality $$ y^2+axy+a(x^3-3xy)+a^3x \geq 2a^2(x^2-y) \, ; $$ the condition $b,c \geq 0$ is equivalent to requiring $x,y \geq 0$.

This inequality is quadratic in $y$. By collecting terms, we see that it suffices to show that \begin{eqnarray} f_{x,a}(y):&=&y^2+y(ax-3ax+2a^2)+ax^3+a^3x-2a^2x^2\\ &=&y^2-2ya(x-a)+ax(x-a)^2 \end{eqnarray} is non-negative whenever $a,x,y$ are.

For fixed $x,a$, the function $f$ is minimized when $y=a(x-a)$. There are two possible cases:

• If $a>x$, this minimum value lies outside the domain of consideration. Since $f$ is a quadratic with positive leading coefficient, it is minimized on the domain of consideration at its boundary; that is, when $y=0$. But $a,x \geq 0$, which means $f_{x,a}(0)=ax(x-a)^2 \geq 0$.
• If $a \leq x$, $f$ is minimized on the domain of consideration at $y=a(x-a)$. Moreover, \begin{eqnarray} f_{x,a}(a(x-a))&=&a^2(x-a)^2-2a^2(x-a)^2+ax(x-a)^2\\ &=& a(x-a)^3 \end{eqnarray} which is non-negative as $a \geq 0$ and $x \geq a$.

So, for any choices of $x$ and $a$, $f$'s minimum value in $y$ is non-negative. Thus $f$ is non-negative on the entire domain of consideration.