# Algebraic Techniques 04 - Lagrange's Identity and Cauchy-Schwarz Inequality

#### ✔ Lagrange’s Identity

\begin{aligned} &\text{For all real numbers} \; x_1, \cdots, x_n, \text{and} \; y_1, \cdots, y_n, \\\\ &\qquad (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2) = (x_1y_1 + \cdots + x_ny_n)^2 + \sum_{1 \leq i < j \leq n} (x_i y_j - x_j y_i)^2 \\\\\\ &\text{Especially, for real numbers} \; a, b, c, d, \\\\ &\qquad (a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2 = (ac - bd)^2 + (ad + bc)^2 \end{aligned}

#### ✔ Cauchy-Schwarz Inequality

\begin{aligned} &\text{For all real numbers} \; x_1, \cdots, x_n, \text{and} \; y_1, \cdots, y_n, \\\\ &\qquad\qquad (x_1^2 + \cdots + x_n^2)(y_1^2 + \cdots + y_n^2) \geq (x_1y_1 + \cdots + x_ny_n)^2 \\\\ &\text{The equal sign is valid when} \\ &\qquad\qquad \text{(1)} \; x_1 = \cdots = x_n = 0, \text{or}\\ &\qquad\qquad \text{(2) there exists a real number} \; t \; \text{such that} \; y_i = t x_i \; (i = 1, \cdots, n) \end{aligned}

#### ✔ Corollary of Cauchy-Schwarz Inequality

\begin{aligned} &\text{For all real numbers} \; a_1, \cdots, a_n, \text{and all positive real numbers} \; b_1, \cdots, b_n, \\\\ &\qquad\qquad\qquad\qquad \frac{a_1^2}{b_1} + \cdots + \frac{a_n^2}{b_n} \geq \frac{(a_1 + \cdots + a_n)^2}{b_1 + \cdots + b_n} \end{aligned}

#### ✔ Minkowski Inequality

\begin{aligned} &\text{For all real numbers} \; x_1, \cdots, x_n, \text{and} \; y_1, \cdots, y_n, \\\\ &\qquad\qquad \sqrt{x_1^2 + y_1^2} + \cdots + \sqrt{x_n^2 + y_n^2} \geq \sqrt{(x_1 + \cdots + x_n)^2 + (y_1 + \cdots + y_n)^2} \end{aligned}

#### For all real numbers $\, a$, $b$, and $c$, prove the following equation.

\begin{aligned} (1 + a^2)(1 + b^2)(1 + c^2) = (a + b + c - abc)^2 + (ab + bc + ca - 1)^2 \end{aligned}

[Solution] Applying Lagrange’s identity twice, \begin{aligned} (1 + a^2)(1 + b^2) &= (a + b)^2 + (1 - ab)^2 \\\\ \Longrightarrow (1 + a^2)(1 + b^2)(1 + c^2) &= ((a + b)^2 + (1 - ab)^2)(1 + c^2) \\ &= (a + b + c(1 - ab))^2 + ((a + b)c - (1 - ab))^2 \\ &= (a + b + c - abc)^2 + (ab + bc + ca - 1)^2 \end{aligned}

#### For real numbers $\, a$ and $b$, prove the following inequality.

\begin{aligned} a^4 + b^4 \geq ab (a^2 + b^2) \end{aligned}

[Solution] By Cauchy-Schwarz Inequality and AM-GM, \begin{aligned} (1^2 + 1^2)(x^2 + y^2) \geq (x + y)^2, \quad \frac{x^2 + y^2}{2} \geq xy \end{aligned}

Therefore, \begin{aligned} a^4 + b^4 \geq \frac{(a^2 + b^2)^2}{2} = \frac{a^2 + b^2}{2} (a^2 + b^2) \geq ab (a^2 + b^2) \end{aligned}

#### For positive real numbers $\, a$, $b$, and $\, c$ such that $\, a^2 + b^2 + c^2 = 3$, prove the following inequality.

\begin{aligned} \frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2} \geq \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \end{aligned}

[Solution] By Cauchy-Schwarz Inequality, \begin{aligned} \left(\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a}\right)^2 = \left(\sqrt{\frac{a^3}{b^2}}\sqrt{a} + \sqrt{\frac{b^3}{c^2}}\sqrt{b} + \sqrt{\frac{c^3}{a^2}}\sqrt{c}\right)^2 \leq \left(\frac{a^3}{b^2} + \frac{b^3}{c^2} + \frac{c^3}{a^2}\right) (a + b + c) \end{aligned}

It is sufficient to prove that \begin{aligned} a + b + c \leq \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \end{aligned}

By corollary of Cauchy-Schwarz Inequality, \begin{aligned} \frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq \frac{(a + b + c)^2}{a + b + c} = a + b + c \end{aligned}

## Reference

[1] Titu Andreescu, 105 Algebra Problems.