Inequality

Monotonous function

Let \(f:A\rightarrow B\).If \(x_1,x_2\in A\) ,such that \(f(x_1)<f(x_2)\),then \(f\) is called a increasing function.
If \(f\) increasing,then \(f'(x)\leq 0\), for \(\forall x\in A\).

• Let \(g\) be a continous function. If \(f\) have an inverse \(f^{-1}\), then \(f,f^{-1}\) are called monotonous functions.
  
• Let \(f:A\rightarrow B\).If for \(\forall x_1,x_2\leq A\). Such that \(x_1<x_2\),then \(f(x_1)<f(x_2)\),then \(f\) is called a strictly increasing function.
  
  • If \(f\) strictly increasing,then \(f'(x)>0\), for \(\forall x\in A\).

\(\clubsuit\) AM-GM inequality

Let \(n\) be a positive integer,let \(a_1,a_2a_3\dots a_n\) be positive real numbers. Then: \[\frac{a_1+a_2+\cdots+a_n}{n}\geq \sqrt[n]{a_1a_2\cdots a_n}\] (Equal when \(a_1=a_2=a_3... =a_n\))

Proof by Strong Induction

• Base Case

  For \(n = 1\), the AM-GM inequality trivially holds since:
  \(\frac{a_1}{1} = a_1 \geq \sqrt[1]{a_1} = a_1\).
  Thus, the base case is satisfied.

• Inductive Hypothesis:

  Assume that the AM-GM inequality holds for all integers \(m\) such that \(1 \leq m \leq k\), i.e., for any set of positive real numbers \(a_1, a_2, \dots, a_m\) with \(m \leq k\), we have:\(\frac{a_1 + a_2 + \dots + a_m}{m} \geq \sqrt[m]{a_1 a_2 \dots a_m}\)

• Inductive Step:

  We need to prove that the AM-GM inequality also holds for \(n = k+1\). That is, for any set of positive real numbers \(a_1, a_2, \dots, a_{k+1}\), we must show:

  \[ \frac{a_1 + a_2 + \dots + a_{k+1}}{k+1} \geq \sqrt[k+1]{a_1 a_2 \dots a_{k+1}} \]

  Consider the set of \(k+1\) positive real numbers \(a_1, a_2, \dots, a_{k+1}\). We can split this set into two subsets:

  • \(a_1, a_2, \dots, a_k\)
  • \(a_{k+1}\)

  Let \(S = \frac{a_1 + a_2 + \dots + a_k}{k}\) denote the arithmetic mean of the first \(K\) numbers.

  By the inductive hypothesis, we know that:

  \[ S \geq \sqrt[k]{a_1 a_2 \dots a_k} \]

  Now, we consider the arithmetic mean and geometric mean of \(S\) and \(a_{k+1}\):

  \[ \frac{S + a_{k+1}}{2} \geq \sqrt{S \cdot a_{k+1}} \]

  Substituting the value of \(S\) and simplifying:

  \[ \frac{\frac{a_1 + a_2 + \dots + a_k}{k} + a_{k+1}}{2} \geq \sqrt{\frac{a_1 a_2 \dots a_k \cdot a_{k+1}}{k}} \]

  Multiplying both sides by \(2\) and rearranging, we obtain:

  \[ \frac{a_1 + a_2 + \dots + a_k + a_{k+1}}{k+1} \geq \sqrt[k+1]{a_1 a_2 \dots a_{k+1}} \]

  Thus, the AM-GM inequality holds for \(n = k+1\).

  By the principle of strong induction, the AM-GM inequality is therefore true for all positive integers \(n\).

Jensen's inequality

\(\frac{d^2{y}}{d{x^2}}\geq 0\)

Sample questions

1. Proof that \(H_n \leq A_n \leq G_n \leq Q_n\)

• \(H_n = \frac{n}{\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}}\)
• \(A_n = \frac{x_1 + x_2 + \dots + x_n}{n}\)
• \(G_n = \sqrt[n]{x_1 \cdot x_2 \cdot \dots \cdot x_n}\)
• \(Q_n = \sqrt{\frac{x_1^2 + x_2^2 + \dots + x_n^2}{n}}\)