Created on August 02, 2023

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Summary: The given text consists of two parts. The first part contains two examples along with their proofs concerning inequalities. In the first example, it is proved that if 'a_i' are real numbers such that the sum of their squares, the summation of (a_i^2), is less than or equal to 4, then the sum of their cubes, the summation of (a_i^3), will be less than or equal to 8. The proof involves utilizing the AM-GM inequality to demonstrate that the sum of the cubes is bounded by 8. In the second example, an inequality is proved for non-negative numbers 'a', 'b', and 'c': a^3 + b^3 + c^3 - 3abc is greater than or equal to 2((b+c)/2 - a)^3.

$\textbf{Example. }$Let $a_i$ be real numbers, for $1 \le i\le n,$ satisfying $\sum_{i=1}^{n} a_i^2 \le 4$, then $\sum_{i=1}^n a_i^3 \le 8$

Proof.

We notice that if $a_i^2 \le 4$, then we will have $|a_i| \le 2$ and so $a_i^2(a_i-2) \le 0 \implies a_i^3 \le 2a_i^2$ and so we will have

$$\sum_{i=1}^n a_i^3 \le 2\sum_{i=1}^na_i^2 \le 8.$$

$\textbf{Example. }$If $a, b, c$ are non-negative numbers, then

$$a^3+b^3+c^3-3 a b c \geq 2\left(\frac{b+c}{2}-a\right)^3$$

Proof.

Notice that $a^3 +b^3 +c^3 - 3abc \ge 0$ by AM-GM inequality and if $\left(\frac{b+c}{2}-a\right) \le 0$, then the original inequality holds.

Now let's consider $\frac{b+c}{2}-a>0,$ which means $b+c >2a$.

Then we will have $$\begin{equation} \begin{aligned} &a^3+b^3+c^3-3 a b c - 2\left(\frac{b+c}{2}-a\right)^3 \\ &= a^3+b^3+c^3-3 a b c - \frac{1}{4}\left(b-a+c-a\right)^3 \\ &= a^3 +(b-a+a)^3 +(c-a+a)^3 -3a(b-a+a)(c-a+a) - \frac{1}{4}(b-a+c-a)^3 \\ &= a^3 +(b-a)^3 +3(b-a)^2 a+3(b-a)a^2 +a^3 \\ &\quad +(c-a)^3 +3(c-a)^2a+3(c-a)a^2+a^3 \\ &\quad +(-3(b-a)(c-a)a-3(b-a)a^2-3(c-a)a^2-3a^3) \\ &\quad -\frac{1}{4}\left((b-a)^3 + 3(b-a)^2(c-a)+3(b-a)(c-a)^2 + (c-a)^3\right) \\ &= \frac{3}{4}(b-a)^3 +3(b-a)^2 a +\frac{3}{4}(c-a)^3 +3(c-a)^2a \\ &\quad +(-3(b-a)(c-a)a) -\frac{1}{4}\left( + 3(b-a)^2(c-a)+3(b-a)(c-a)^2 \right) \\ &= \frac{3}{4}(b-a)^3 +3(b-a)^2 a +\frac{3}{4}(c-a)^3 +3(c-a)^2a \\ &\quad +(-3(b-a)(c-a)a) -\frac{3}{4}\left( (b-a)+(c-a) \right)(b-a)(c-a) \\ &= \frac{3}{4} \times((b-a)^3 +(c-a)^3 +(b-a+c-a)(b-a)(c-a)) \\ &\quad +3a((b-a)^2+(c-a)^2-(b-a)(c-a)) \\ &\ge \frac{3}{4} \times ((b-a)^3 +(c-a)^3 +(b-a+c-a)(b-a)(c-a)) \\ &= \frac{3}{4} \times ((b-a+c-a)\times ((b-a)^2 - (b-a)(c-a)+(c-a)^2)+(b-a+c-a)(b-a)(c-a)) \\ &= \frac{3}{4} \times(b-a+c-a) \times((b-a)^2+(c-a)^2) \ge 0 \end{aligned} \end{equation}$$

This finishes the prove.

Food for thoughts

Prove or give counter example:

$$\sum_{i=1}^n a_i^n - n\prod_{i=1}^na_i\ge (n-1) \left(\frac{\sum_{i=2}^n a_i }{n-1}-a_1\right)^3 $$