how do you prove $9(a^3+b^3+c^3)$ $\ge$ $(a+b+c)^3$
I tried to expand by multinomial expansion the right side and got a long string so what do i do next?
Asked
Active
Viewed 83 times
1 Answers
0
By a special case of Hölder's inequality we have
$$\sum_{k=1}^n |x_ky_k| \leq \left(\sum_{k=1}^n x_k^p\right)^\frac{1}{p} \left(\sum_{k=1}^n y_k^q\right)^\frac{1}{q}$$
Where $1/p + 1/q = 1$.
Let $n=3$, $x_1 = a$, $x_2 = b$ and $x_3 = c$ and $y_1 = y_2 = y_3 = 1$. Let also $p = 3$ and $q = 3/2$. Then:
$$\frac{|a| + |b| + |c|}{3} \leq \sqrt[3]{\frac{a^3 + b^3 + c^3}{27}} $$
Equivalently,
$$(|a|+|b|+|c|)^3 \leq 9(a^3 + b^3 + c^3)$$
And then it follows that
$$(a+b+c)^3 \leq 9(a^3 + b^3 + c^3)$$
Darth Geek
- 12,296