Proofs of Two Inequalities Involving Positive Integers and Non-negative Numbers

Show that for positive integers $abc=1$, we have $$\frac{a+b+c}{3} \ge \sqrt[5]{\frac{a^2+b^2+c^2}{3}}$$

Proof.

Since we have

$$\frac{a+b+c}{3} \ge \sqrt[5]{\frac{a^2+b^2+c^2}{3}} \\ \implies \left(\frac{a+b+c}{3}\right)^5 \ge \frac{a^2 +b^2+c^2}{3} \\ \implies (a+b+c)^5\ge 81 (a^2+b^2+c^2)$$

If we homogenize the equation, we will have

$$(a+b+c)^5 \ge 81 abc(a^2+b^2+c^2)$$

Since $abc =1$, we can further assume $a+b+c=3$, then

we want to show $f(a,b,c) = abc(a^2+b^2+c^2)$ obtain its maximal at $a=b=c=1$.

Assume that $f$ obtains its maximum at $(a,b,c)$ and $a \not= b$. Then we can construct $f(\frac{a+b}{2}, \frac{a+b}{2}, c)$ has bigger value than $f(a,b,c)$

$$f\left(\frac{a+b}{2}, \frac{a+b}{2}, c\right) - f(a,b,c) = \left(\frac{a+b}{2}\right)^2 c\left(2\left(\frac{a+b}{2}\right)^2+c^2\right)-abc(a^2+b^2+c^2) \\ = c^3 \left[\left(\frac{a+b}{2}\right)^2-ab\right] + \left[2 \left(\frac{a+b}{2}\right)^4-(a^2+b^2)ab\right]c \\=c^3 \frac{(a-b)^2}{4} + \frac{(a-b)^4}{8}c>0$$

so $(\frac{a+b}{2}, \frac{a+b}{2}, c)$ is bigger, which is a contradiction.

Therefore $a=b=c=1$ is the maximum and the equality obtained in this situation.

Inequality 2.

$$a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0$$

for non-negative numbers $a,b,c \in \mathbb{R}$ and $r >0$.

Proof.

Since $a,b,c$ are all symmetric, without loss of generality, let's assume that $a\ge b\ge c \ge 0$, then we will have

$$ LHS = (a-b)\left(a^r (a-c) -b^r(b-c)\right)+c^r (c-a)(b-a) $$

Notice that $a-c \ge b-c \ge 0 $ and $a^r \ge b^r$ so we will have $a^r (a-c) \ge b^r (b-c) \implies a^r (a-c) - b^r (b-c) \ge 0 $ and we also have $c-a \le 0$ and $b-a\le0$ so overall we have

$$(a-b)\left(a^r (a-c) -b^r(b-c)\right)+c^r (c-a)(b-a) \ge 0 $$