# Algebraic Techniques 03 - Inequalities

#### ✔ Quadratic Mean(RMS), Arithmetic Mean, Geometric Mean, and Harmonic Mean

\begin{aligned} \text{For all positive real numbers} \; a_1, \cdots, a_n&, \\\\ \sqrt{\frac{a_1^2 + \cdots + a_n^2}{n}} \geq \frac{a_1 + \cdots + a_n}{n} &\geq \sqrt[n]{a_1 \cdots a_n} \geq \frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}} \end{aligned}

#### ✔ Hölder’s Inequality

\begin{aligned} \text{For all non-negative real numbers} \; a_{11}, \cdots, &a_{1n}, a_{k1}, \cdots, a_{kn}, \\\\ (a_{11} + \cdots + a_{1n})\cdots(a_{k1} + \cdots + a_{kn}) &\geq \left( \sqrt[k]{a_{11} \cdots a_{1n}} + \cdots + \sqrt[k]{a_{k1} \cdots a_{kn}} \right)^k \end{aligned}

#### For positive real numbers $\, x$ and $y$, prove the following inequality.

\begin{aligned} \frac{x}{x^4 + y^2} + \frac{y}{y^4 + x^2} \leq \frac{1}{xy} \end{aligned}

[Solution] By AM-GM on only denominators, \begin{aligned} \frac{x}{x^4 + y^2} \leq \frac{x}{2x^2y} = \frac{1}{2xy}, \quad \frac{y}{y^4 + x^2} \leq \frac{y}{2y^2x} = \frac{1}{2xy} \end{aligned}

The problem is proved after adding two inequalities.

#### For real numbers $\, a$, $b$, and $c$ in $(0, 4)$, prove that at least one of the following inequalities is $\geq 1$.

\begin{aligned} \frac{1}{a} + \frac{1}{4 - b}, \quad \frac{1}{b} + \frac{1}{4 - c}, \quad \frac{1}{c} + \frac{1}{4 - a} \end{aligned}

[Solution] For a proof by contradiction, let the statement negate. That is, assume that \begin{aligned} \frac{1}{a} + \frac{1}{4 - b} < 1, \quad \frac{1}{b} &+ \frac{1}{4 - c} < 1, \quad \frac{1}{c} + \frac{1}{4 - a} < 1 \\\\ \Longrightarrow \frac{1}{a} + \frac{1}{4 - a} + \frac{1}{b} &+ \frac{1}{4 - b} + \frac{1}{c} + \frac{1}{4 - c} < 3 \end{aligned}

By AM-HM, \begin{aligned} \frac{1}{a} + \frac{1}{4 - a} &\geq \frac{2^2}{a + (4 - a)} = 1 \\\\ \Longrightarrow \frac{1}{a} + \frac{1}{4 - a} + \frac{1}{b} &+ \frac{1}{4 - b} + \frac{1}{c} + \frac{1}{4 - c} \geq 3 \end{aligned}

which means the contradiction.

#### Given the following inequalities for positive real numbers $\, a$ and $b$, prove that $\, a + b \leq 2$.

\begin{aligned} |a - 2b| \leq \frac{1}{\sqrt{a}} \quad \text{and} \quad |b - 2a| \leq \frac{1}{\sqrt{b}} \end{aligned}

[Solution] After multiplying each inequality by $\sqrt{a}$ and $\sqrt{b}$ and square them, \begin{aligned} a(a - 2b)^2 \leq 1, \quad b(2a - b)^2 \leq 1 \end{aligned}

Adding the above inequalities, \begin{aligned} a^3 - 4a^2b + 4ab^2 + 4a^2b - 4ab^2 + b^3 = a^3 + b^3 \leq 2 \end{aligned}

By Hölder’s inequality, \begin{aligned} 8 &\geq 4(a^3 + b^3) = (1^3 + 1^3)(1^3 + 1^3)(a^3 + b^3) \geq (a + b)^3 \\\\ \Longrightarrow \; &a + b \leq 2 \end{aligned}

#### Given the following inequality for positive real numbers $\, x$, $y$, and $z$, prove that $\, x + y + z \geq \sqrt{3}$.

\begin{aligned} xy + yz + zx \geq \frac{1}{\sqrt{x^2 + y^2 + z^2}} \end{aligned}

[Solution] Simplifying the condition more which is very tricky, \begin{aligned} (xy + yz + zx)^2 (x^2 + y^2 + z^2) \geq 1 \end{aligned}

By AM-GM, \begin{aligned} (xy + yz + zx)^2 (x^2 + y^2 + z^2) \leq \left( \frac{2(xy + yz + zx) + x^2 + y^2 + z^2}{3} \right)^3 = \left( \frac{(x + y + z)^2}{3} \right)^3 \end{aligned}

It implies that \begin{aligned} \left( \frac{(x + y + z)^2}{3} \right)^3 \geq 1 \Longleftrightarrow \left( \frac{(x + y + z)^2}{3} \right) \geq 1 \Longleftrightarrow x + y + z \geq \sqrt{3} \end{aligned}

## Reference

[1] Titu Andreescu, 105 Algebra Problems.