Algebraic Techniques 03 - Inequalities

• ✔ Quadratic Mean(RMS), Arithmetic Mean, Geometric Mean, and Harmonic Mean
• ✔ Hölder’s Inequality
• ✔ Jensen’s Inequality
• For positive real numbers $\, x$ and $y$, prove the following inequality.
• For real numbers $\, a$, $b$, and $c$ in $(0, 4)$, prove that at least one of the following inequalities is $\geq 1$.
• Given the following inequalities for positive real numbers $\, a$ and $b$, prove that $\, a + b \leq 2$.
• Given the following inequality for positive real numbers $\, x$, $y$, and $z$, prove that $\, x + y + z \geq \sqrt{3}$.
• Given $\, A$, $B$, and $C$ are the three angles of a triangle, find the minimum of
• For real numbers $\, a$, $b$, and $c$, prove the following inequality.
• References

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

$$\begin{aligned} \text{For all positive real numbers} \; a_1, \cdots, a_n&, \\\\ \sqrt{\frac{a_1^2 + \cdots + a_n^2}{n}} \geq \frac{a_1 + \cdots + a_n}{n} &\geq \sqrt[n]{a_1 \cdots a_n} \geq \frac{n}{\frac{1}{a_1} + \cdots + \frac{1}{a_n}} \\\\ \text{The equality holds if and only if} \; a_1 = \cdots = a_n. \end{aligned}$$

The condition that this equality holds is often quite useful when finding a minimum.

✔ Hölder’s Inequality

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

✔ Jensen’s Inequality

$$\begin{aligned} \text{For all on-negative real numbers} \; \lambda_{1}, \cdots, \lambda_{n} \; \text{such that} \; \lambda_{1} + \cdots + \lambda_{n} = 1, \\\\ \begin{cases} \lambda_{1} f(x_1) + \cdots + \lambda_{n} f(x_n) \geq f(\lambda_{1} x_1 + \cdots + \lambda_{n} x_n) \quad \text{(} f \text{ is convex)} \\\\ \lambda_{1} f(x_1) + \cdots + \lambda_{n} f(x_n) \leq f(\lambda_{1} x_1 + \cdots + \lambda_{n} x_n) \quad \text{(} f \text{ is concave)} \end{cases} \end{aligned}$$

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

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

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[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 $\geq 1$.

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

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[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 $\, a + b \leq 2$.

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

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[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 $\, x + y + z \geq \sqrt{3}$.

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

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

Given $\, A$, $B$, and $C$ are the three angles of a triangle, find the minimum of

$$\begin{aligned} \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} \end{aligned}$$

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[Solution] Since $A + B + C = \pi$, each $\cot$ above is positive. By AM-GM, $$\begin{aligned} \cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2} \geq 3 \sqrt[3]{\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}} \end{aligned}$$

Recalling that the equality holds if and only if $A/2 = B/2 = C/2$, the minimum of this expression can be found when they are all $\pi / 6$. So, the minimum is $$\begin{aligned} 3 \sqrt[3]{\cot \frac{\pi}{6} \cot \frac{\pi}{6} \cot \frac{\pi}{6}} = 3 \sqrt[3]{\sqrt{3}^3} = 3 \sqrt{3} \end{aligned}$$

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

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

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[Solution] Let $f(x) = x^2$. Since $f$ is convex, by Jensen’s inequality, $$\begin{aligned} \frac{1}{3} a^2 + \frac{1}{3} b^2 + \frac{1}{3} c^2 &\geq \left( \frac{a + b + c}{3} \right)^2 \\\\ \Longrightarrow a^2 + b^2 + c^2 &\geq \frac{(a + b + c)^2}{3} \end{aligned}$$

Note that $\lambda_{1} = \lambda_{2} = \lambda_{3} = 1/3$ and $\lambda_{1} + \lambda_{2} + \lambda_{3} = 1$.

References

[1] Titu Andreescu, 105 Algebra Problems.

[2] IMO기출문제풀이집