Lemma. If $a, b \geq 0$, then $a^3 +b^3 \geq a^2 b + a b ^ 2$. Proof. Since $a,b\geq 0$, $a+b\geq 0$. Also, by the trivial inequality, $(a-b)^2 \geq 0$. Thus, $(a-b)^2 (a+b)\geq 0$. Expanding, $a^3 + b^3 – a^2 b – a b^2 \geq 0$. So, $a^3 +…
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Math Formulas
$(a-b)^3=a^3-3a^2 b+3a b^2-b^3$ (remember the 3’s) $(a+b)^3=a^3+3a^2 b+3a b^2+b^3$ (remember the 3’s) $1^2 + 2^2 + … + n^2 = \frac{n(n+1)(2n+1)}{6}$ $1^3 + 2^3 + … + n^3 = \frac{n^2 (n+1)^2}{4}$ $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ For an $m \times n$ grid of numbers satisfying: let $a_{i,j}$ = the entry in…