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Algebra Difficulty 6.9 National olympiad Prove it

3.395. Prove that if α+β=60(α>0,β>0)\alpha+\beta=60^{\circ}(\alpha>0, \beta>0), then tgαtgβ13\operatorname{tg} \alpha \cdot \operatorname{tg} \beta \leq \frac{1}{3}.

Solution

## Solution.

tgαtgβ13sinαsinβcosαcosβ1312(cos(αβ)cos(α+β))12(cos(αβ)+cos(α+β))13 \begin{aligned} & \operatorname{tg} \alpha \cdot \operatorname{tg} \beta \leq \frac{1}{3} \Leftrightarrow \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} \leq \frac{1}{3} \Leftrightarrow \\ & \Leftrightarrow \frac{\frac{1}{2}(\cos (\alpha-\beta)-\cos (\alpha+\beta))}{\frac{1}{2}(\cos (\alpha-\beta)+\cos (\alpha+\beta))} \leq \frac{1}{3} \end{aligned}

(cos(αβ)cos(α+β))(cos(αβ)+cos(α+β))13 \frac{(\cos (\alpha-\beta)-\cos (\alpha+\beta))}{(\cos (\alpha-\beta)+\cos (\alpha+\beta))} \leq \frac{1}{3}

Since α+β=60\alpha+\beta=60^{\circ}, the inequality becomes

2cos(αβ)12cos(αβ)+1134cos(αβ)4,cos(αβ)1(cos(αβ)0) \frac{2 \cos (\alpha-\beta)-1}{2 \cos (\alpha-\beta)+1} \leq \frac{1}{3} \Leftrightarrow 4 \cos (\alpha-\beta) \leq 4, \cos (\alpha-\beta) \leq 1(\cos (\alpha-\beta) \geq 0)

The last inequality is true. This completes the proof.

## PROGRESSIONS

## BASIC CONCEPTS AND FORMULAS

## Arithmetic Progression

An arithmetic progression is a sequence in which the first term a1a_{1} is given, and each subsequent term, starting from the second, is equal to the previous term plus a constant number dd, called the common difference of the progression.

If the first term a1a_{1} and the common difference dd of an arithmetic progression are given, the nn-th term of the arithmetic progression is calculated by the formula

an=a1+d(n1). a_{n}=a_{1}+d(n-1) .

Formula (4.1) is called the general term formula of an arithmetic progression.

## Properties of the Terms of an Arithmetic Progression

1. Each middle term of an arithmetic progression is equal to the half-sum of the terms equally distant from it:

ak=ak1+ak+12,k=2,3,n1 a_{k}=\frac{a_{k-1}+a_{k+1}}{2}, k=2,3 \ldots, n-1

2. In a finite arithmetic progression, the sums of terms equally distant from the ends are equal to each other and equal to the sum of the extreme terms:

a1+an=a2+an1=ak+ank+1==2a1+d(n1) a_{1}+a_{n}=a_{2}+a_{n-1}=a_{k}+a_{n-k+1}=\ldots=2 a_{1}+d(n-1)

## Sum of the First nn Terms of an Arithmetic Progression

The sum of the first nn terms of an arithmetic progression is equal to

Sn=a1+an2n S_{n}=\frac{a_{1}+a_{n}}{2} \cdot n

Considering (4.3), i.e., that a1+an=2a1+d(n1)a_{1}+a_{n}=2 a_{1}+d(n-1), formula (4.4) can be written as

Sn=2a1+d(n1)2n S_{n}=\frac{2 a_{1}+d(n-1)}{2} \cdot n

## Geometric Progression

A geometric progression is a sequence in which the first term b1b_{1} is given, and each subsequent term, starting from the second, is equal to the previous term multiplied by a constant number qq, called the common ratio of the progression.

If the first term b1b_{1} and the common ratio qq of a geometric progression are given, the nn-th term of the geometric progression is calculated by the formula

bn=b1qn1 b_{n}=b_{1} q^{n-1}

Formula (4.6) is called the general term formula of a geometric progression.

## Properties of the Terms of a Geometric Progression

1. The square of each middle term of a geometric progression is equal to the product of the terms equally distant from it, i.e.,

bk2=bk1bk+1,k=2,3,n1 b_{k}^{2}=b_{k-1} b_{k+1}, k=2,3 \ldots, n-1

2. In a finite geometric progression, the products of terms equally distant from the ends are equal to each other and equal to the product of the extreme terms:

b1bn=b2bn1=b3bn2==bkbnk+1==b12qn1 b_{1} \cdot b_{n}=b_{2} \cdot b_{n-1}=b_{3} \cdot b_{n-2}=\ldots=b_{k} \cdot b_{n-k+1}=\ldots=b_{1}^{2} \cdot q^{n-1}

3. The product of the first nn terms of a geometric progression with positive terms is equal to the nn-th root of the product of its extreme terms:

Pn=(b1bn)n P_{n}=\sqrt{\left(b_{1} \cdot b_{n}\right)^{n}}

In the general case,

Pn=b1bnn \left|P_{n}\right|=\sqrt{\left|b_{1} \cdot b_{n}\right|^{n}}

## Sum of the First nn Terms of a Geometric Progression

The sum of the first nn terms of a geometric progression is calculated by the formula

Sn=b1bnq1q(q1) S_{n}=\frac{b_{1}-b_{n} q}{1-q}(q \neq 1)

Considering (4.6), i.e., that bn=b1qn1b_{n}=b_{1} q^{n-1}, formula (4.10) can be written as

Sn=b1(1qn)1q S_{n}=\frac{b_{1}\left(1-q^{n}\right)}{1-q}

## Sum of the Terms of an Infinite Geometric Progression

The infinite numerical series formed by the terms of a geometric progression b1+b2+b3++bn+b_{1}+b_{2}+b_{3}+\ldots+b_{n}+\ldots converges when q<1|q|<1, and its sum SS is equal to

S=b11q S=\frac{b_{1}}{1-q}

Formula (4.12) is also called the formula for the sum of the terms of an infinitely decreasing geometric progression.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.