Inequalities

Proof. Let \(x\in\mathbb{R}\) with \(x > -1\). Let \(A = \{n\in\mathbb{N} : (1+x)^n \geq 1+nx\}\). Since \((1+x)^0 = 1 = 1+0x\), we have \(0\in A\). Suppose \(k\in A\). Then \((1+x)^k \geq 1+kx\). Since \(x > -1\), we have \(1+x > 0\). Then \[(1+x)^{k+1} = (1+x)^k (1+x) \geq (1+kx)(1+x) = 1 + (k+1)x + kx^2 \geq 1 + (k+1)x\] and it follows that \(k+1\in A\). By induction, \(A=\mathbb{N}\), completing the proof. ◻

Proof. Let \((a_1,\dots,a_n), (b_1,\dots,b_n)\in\mathbb{R}^n\). Let \[A = \sqrt{\sum_{k=1}^n a_k^2}, \quad B = \sqrt{\sum_{k=1}^n b_k^2}, \quad\text{and}\quad C = \sum_{k=1}^n a_k b_k\] so that the inequality to be proven is equivalent to \(C^2\leq A^2B^2\). We consider three cases.

Case 1: Suppose \(A=0\). Then \(a_k=0\) for each \(k=1,\dots,n\). Hence \(C=0\) and \(C^2=A^2B^2=0\).

Case 2: Suppose \(B=0\). Then \(b_k=0\) for each \(k=1,\dots,n\). Hence \(C=0\) and \(C^2=A^2B^2=0\).

Case 3: Suppose \(A>0\) and \(B>0\). Then \[\left|\frac{C}{AB}\right| = \left|\sum_{k=1}^n \frac{a_k b_k}{AB} \right| \leq \sum_{k=1}^n \left| \frac{a_k}{A} \cdot \frac{b_k}{B} \right| \leq \sum_{k=1}^n \frac{1}{2} \left( \frac{a_k^2}{A^2} + \frac{b_k^2}{B^2} \right) = \frac{1}{2} \left( \frac{\sum_{k=1}^n a_k^2}{A^2} + \frac{\sum_{k=1}^n b_k^2}{B^2} \right) = 1\] so that \(|C| \leq |AB|\). Then \(C^2\leq A^2B^2\).

If there is \(c\in\mathbb{R}\) such that \(b_k = ca_k\) for each \(k=1,\dots,n\), then \[C^2 = \left( \sum_{k=1}^n a_k b_k \right)^2 = \left( \sum_{k=1}^n c a_k^2 \right)^2 = \left( \sum_{k=1}^n a_k^2 \right) \cdot \left( \sum_{k=1}^n c^2 a_k^2 \right) = \left(\sum_{k=1}^n a_k^2\right) \cdot \left(\sum_{k=1}^n b_k^2\right) = A^2B^2.\] Suppose \(C^2=A^2B^2\). Let \(c = B/A\). WIP ◻