0.1 Advanced Inequalities
Muirhead’s inequality is one of the powerful proving tools that is also a generalization of AM-GM inequality.
Theorem 0.1.1: Muirhead’s Inequality
Let \((a_i)\) and \((b_i)\) be sequences such that \((a_i) \succ (b_i)\). Then, the following inequality holds for all nonnegative \(x_i\). \[ \sum _\text {sym} {x_1}^{a_1} \cdots {x_n}^{a_n} \geq \sum _\text {sym} {x_1}^{b_1} \cdots {x_n}^{b_n} \] The equality holds if and only if \((a_i)\) and \((b_i)\) are identical or if all \(x_i\) are identical, which is the most common case.
Another very powerful inequality is Shur’s inequality. Shur’s inequality comes with many alternative forms, and it is generally encouraged to prove the derivation as there are also not-so-famous ones.
Theorem 0.1.2: Schur’s Inequality
For \(a, b, c \geq 0\) and \(r \geq 0\), \begin{align*} \sum _\text {cyclic} a^r(a - b)(a - c) &\geq 0 \\ a^r(a - b)(a - c) + b^r(b - a)(b - c) + c^r(c - a)(c - b) &\geq 0 \end{align*}
holds with equality if \(a = b = c\).
Proof.
First, notice that the left hand side of the inequality is cyclic. Therefore without loss of generality, let \(a \geq b \geq c \geq 0\). Rewriting the inequality, \begin{align*} a^r(a - b)(a - c) + b^r(b - a)(b - c) + c^r(c - a)(c - b) &\geq 0 \\ (a - b)\left [ a^2(a - c) - b^r(b - c) \right ] + c^r(c - a)(c - b) &\geq 0 \end{align*}
holds. Moreover by construction, \(a - b\), \(a^2(a - c) - b^r(b - c)\), and \(c^r(c - a)(c - b)\) are all greater than or equal to zero. Since the sum of all non-negative numbers is a non-negative number, the inequality holds.
It won’t be an exaggeration to say that the real power of Schur’s inequality in olympiads come from its corollaries.
Corollary 0.1.3
The following inequality holds for all reals \(a, b, c\). \[ a^2 + b^2 + c^2 \geq ab + bc + ca \]
Proof.
Substituting \(r = 0\) in Schur’s inequality, the following is obtained. \begin{align*} (a - b)(a - c) + (b - a)&(b - c) + (c - a)(c - b) \geq 0 \\ \therefore a^2 + b^2 + c^2 &\geq ab + bc + ca \end{align*}
However, notice that unlike the condition for \(a, b, c\) in Schur’s inequality, the inequality above holds for negative \(a, b, c\). Without loss of generality, let \(a \geq b \geq c\). If \(c < 0\) and \(a, b \geq 0\), then it is self evident that the inequality is holds. Similarly, if two variables are negative, the inequality also holds. Lastly, when all \(a, b, c\) are all negative, then the inequality holds as positive numbers are greater than negative sum.
Minkowski’s inequality is a generalization of the triangle inequality.
Theorem 0.1.4: Minkowski’s Inequality
For \(a_i, b_i > 0\) and \(p > 1\), the following inequality is holds. \[ \left ( \sum _{k = 1}^n (a_k + b_k)^p \right )^\frac {1}{p} \leq \left ( \sum _{k = 1}^n a_k^p \right )^\frac {1}{p} + \left ( \sum _{k = 1}^n b_k^p \right )^\frac {1}{p} \] The equality holds if and only if there exists \(\lambda > 0\) such that \(a_i = \lambda b_i\) for all \(1 \leq i \leq n\).
* Bernoulli’s Inequality * Equal Value Principle
0.1.1 Symmetric and Cyclic Inequalities
* Newton’s Inequality * Maclaurin’s Inequality * Popoviciu’s Inequality
Nesbitt’s inequality can be considered as the direct consequence of Cauchy inequality.
Theorem 0.1.5: Nesbitt’s Inequality
For arbitrary \(a, b, c > 0\), the following inequality holds. \[ \sum _\text {cyc} \frac {a}{b + c} \geq \frac {3}{2} \]
Proof.
Consider the following inequality by Cauchy inequality. \begin{align*} \left ( (a + b) + (b + c) + (c + a) \right ) &\left ( \frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a} \right ) \\ &\qquad \qquad \qquad \quad \geq (1 + 1 + 1)^2 = 9 \end{align*}
Therefore, the following inequalities hold. \begin{align*} &\quad \,\ 2 \left ( \frac {a + b + c}{a + b} + \frac {a + b + c}{b + c} + \frac {a + b + c}{c + a} \right ) \\ &= 2 \left ( \frac {c}{a + b} + \frac {a}{b + c} + \frac {b}{c + a} + 3 \right ) \geq 9 \end{align*}
Thus, the inequality holds.
Nesbitt’s inequality is less frequently used than other inequalities, but when applied, it can be a very powerful inequality.