Confusions in Holder's Inequality

Question

Holder's Inequality states that for nonnegative real numbers $a_1,...,a_n$ and $b_1,...,b_n$ we have $$\left(\sum_{i=1}^na_i\right)^p\left(\sum_{i=1}^nb_i\right)^q\ge \left(\sum_{i=1}\sqrt[p+q]{a_i^pb_i^q}\right)^{p+q}$$ Where $p$ and $q$ are positive real numbers. Here is my problem : $a,b,c$ are positive reals, prove that $$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge (a+b+c)^3$$

But for $x>0$ we have $x^5-x^2+3\ge x^3+2$ so we only need to prove $$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3$$ But when I read the solution, it says ''From Holder's Inequality, it follows '' $$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3$$

But I don't see that.

Answer 1

It should be $a^5 - a^2 + 3$ etc. rather than $a^5 - a^2 + 2$ etc.

Problem: Let $a, b, c $ be positive reals. Prove that $$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge (a+b+c)^3.$$

Proof: It is easy to prove that $x^5 - x^2 + 3 \ge x^3 + 2$ for all $x \ge 0$.

It suffices to prove that $$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3.$$

Using Holder's inequality, we have \begin{align*} (a^3+2)(b^3+2)(c^3+2) &= (a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3)\\ &\ge (\sqrt[3]{a^3 \cdot 1 \cdot 1} + \sqrt[3]{1 \cdot b^3 \cdot 1} + \sqrt[3]{1\cdot 1 \cdot c^3})^3\\ & = (a + b + c)^3. \end{align*}

For Holder's inequality, see: https://artofproblemsolving.com/wiki/index.php/H%C3%B6lder%27s_Inequality

Answer 2

I would think Holder's inequality implies

$$(a^3+2)(b^3+2)(c^3+2)\geq \left((a^3b^3c^3)^{1/3}+ (2^3)^{1/3} \right)^3=(abc+2)^3$$

So it would be good to know how your last inequality follows from Holder's.