Prove A ( B + C ) B 2 + B C + 4 C 2 + B ( C + A ) C 2 + C A + 4 A 2 + C ( A + B ) A 2 + A B + 4 B 2 ≥ 1. \frac{a(b+c)}{b^2+bc+4c^2}+\frac{b(c+a)}{c^2+ca+4a^2}+\frac{c(a+b)}{a^2+ab+4b^2}\ge 1. B 2 + B C + 4 C 2 A ( B + C ) + C 2 + C A + 4 A 2 B ( C + A ) + A 2 + Ab + 4 B 2 C ( A + B ) ≥ 1.

Mar 8, 2025 by ADMIN 296 views

$Prove $\frac{a(b+c)}{b^2+bc+4c^2}+\frac{b(c+a)}{c^2+ca+4a^2}+\frac{c(a+b)}{a^2+ab+4b^2}\ge 1.$$

Introduction

In this article, we will explore a challenging inequality involving three variables $a$ , $b$ , and $c$ . The given inequality is $\frac{a(b+c)}{b^2+bc+4c^2}+\frac{b(c+a)}{c^2+ca+4a^2}+\frac{c(a+b)}{a^2+ab+4b^2}\ge 1.$ We will examine various approaches to solve this inequality, including the Cauchy-Schwarz inequality, the Arithmetic Mean-Geometric Mean (AM-GM) inequality, and other classical methods.

Problem Statement

Let $a,b,c\ge 0$ and consider the inequality $\frac{a(b+c)}{b^2+bc+4c^2}+\frac{b(c+a)}{c^2+ca+4a^2}+\frac{c(a+b)}{a^2+ab+4b^2}\ge 1.$ Our goal is to prove this inequality using a combination of mathematical techniques.

Approach 1: Cauchy-Schwarz Inequality

One possible approach to solving this inequality is to use the Cauchy-Schwarz inequality. This inequality states that for any vectors $\mathbf{x}$ and $\mathbf{y}$ in an inner product space, we have $\left(\sum_{i=1}^{n} x_i y_i\right)^2 \leq \left(\sum_{i=1}^{n} x_i^2\right) \left(\sum_{i=1}^{n} y_i^2\right)$ .

To apply the Cauchy-Schwarz inequality to our problem, we can rewrite the given inequality as follows:

\frac{a(b+c)}{b^2+bc+4c^2}+\frac{b(c+a)}{c^2+ca+4a^2}+\frac{c(a+b)}{a^2+ab+4b^2}\ge 1

\Rightarrow \frac{a^2(b+c)}{b^2+bc+4c^2}+\frac{b^2(c+a)}{c^2+ca+4a^2}+\frac{c^2(a+b)}{a^2+ab+4b^2}\ge a+b+c

Now, we can apply the Cauchy-Schwarz inequality to the left-hand side of the inequality:

\left(\frac{a^2(b+c)}{b^2+bc+4c^2}+\frac{b^2(c+a)}{c^2+ca+4a^2}+\frac{c^2(a+b)}{a^2+ab+4b^2}\right)^2 \leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(b+c)^2}{b^2+bc+4c^2}+\frac{(c+a)^2}{c^2+ca+4a^2}+\frac{(a+b)^2}{a^2+ab+4b^2}\right)

Simplifying the right-hand side of the inequality, we get:

\left(\frac{a^2(b+c)}{b^2+bc+4c^2}+\frac{b^2(c+a)}{c^2+ca+4a^2}+\frac{c^2(a+b)}{a^2+ab+4b^2}\right)^2 \leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

Now, we can use the fact that $a,b,c\ge 0$ to simplify the right-hand side of the inequality:

\left(\frac{a^2(b+c)}{b^2+bc+4c^2}+\frac{b^2(c+a)}{c^2+ca+4a^2}+\frac{c^2(a+b)}{a^2+ab+4b^2}\right)^2 \leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

\leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

\leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

\leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

\leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

\leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

\leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left(\frac{(a+b+c)^2}{a^2+ab+4b^2}+\frac{(a+b+c)^2}{c^2+ca+4a^2}+\frac{(a+b+c)^2}{b^2+bc+4c^2}\right)

\leq \left(\frac{a^2}{b^2+bc+4c^2}+\frac{b^2}{c^2+ca+4a^2}+\frac{c^2}{a^2+ab+4b^2}\right) \left<br/> # Q&A: Prove $\frac{a(b+c)}{b^2+bc+4c^2}+\frac{b(c+a)}{c^2+ca+4a^2}+\frac{c(a+b)}{a^2+ab+4b^2}\ge 1.$

Q: What is the given inequality and what are the constraints on the variables $a$ , $b$ , and $c$ ?

A: The given inequality is $\frac{a(b+c)}{b^2+bc+4c^2}+\frac{b(c+a)}{c^2+ca+4a^2}+\frac{c(a+b)}{a^2+ab+4b^2}\ge 1.$ The constraints on the variables $a$ , $b$ , and $c$ are that they are non-negative, i.e., $a,b,c\ge 0$ .

Q: What is the approach to solving this inequality?

A: One possible approach to solving this inequality is to use the Cauchy-Schwarz inequality. This involves rewriting the given inequality in a form that allows us to apply the Cauchy-Schwarz inequality, and then simplifying the resulting expression to obtain the desired inequality.

Q: How do we apply the Cauchy-Schwarz inequality to the given inequality?

A: To apply the Cauchy-Schwarz inequality, we first rewrite the given inequality as follows: