How To Prove An Inequality With $a+b+c=3$?

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Introduction

In this article, we will explore a challenging inequality problem involving symmetric polynomials and the AM-GM inequality. The problem statement is as follows: given that $a, b, c \ge 0$ and $a + b + c = 3$, we need to prove the inequality:

$$\frac{a}{2a^{2}-5a+15}+\frac{b}{2b^{2}-5b+15}+\frac{c}{2c^{2}-5c+15}\le \frac{1}{4}.$$

Background and Motivation

The given inequality involves symmetric polynomials, which are expressions that remain unchanged under any permutation of the variables. In this case, we have a cubic polynomial in each denominator, which can be factored as:

$$2a^{2}-5a+15 = (a-3)(2a-5).$$

This factorization suggests that we may be able to use the AM-GM inequality to establish a lower bound for each term in the sum.

Using the AM-GM Inequality

The AM-GM inequality states that for any non-negative real numbers $x_1, x_2, \ldots, x_n$, we have:

$$\frac{x_1 + x_2 + \cdots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdots x_n}.$$

We can use this inequality to establish a lower bound for each term in the sum. Specifically, we have:

$$\frac{a}{2a^{2}-5a+15} = \frac{a}{(a-3)(2a-5)} \ge \frac{a}{2\sqrt{(a-3)(2a-5)}}.$$

Establishing a Lower Bound for Each Term

Using the AM-GM inequality, we can establish a lower bound for each term in the sum. Specifically, we have:

$$\frac{a}{2a^{2}-5a+15} \ge \frac{a}{2\sqrt{(a-3)(2a-5)}} \ge \frac{a}{2\sqrt{2(a-3)(a-5/2)}}.$$

Simplifying the Expression

We can simplify the expression further by using the fact that $a + b + c = 3$. Specifically, we have:

$$\frac{a}{2a^{2}-5a+15} \ge \frac{a}{2\sqrt{2(a-3)(a-5/2)}} = \frac{a}{2\sqrt{2(a-3)(a-5/2)}} \cdot \frac{a+3}{a+3}.$$

Canceling Terms

We can cancel terms in the numerator and denominator to obtain:

$$\frac{a}{2a^{2}-5a+15} \ge \frac{a(a+3)}{2(a-3)(a-5/2)(a+3)}.$$

Factoring the Denominator

We can factor the denominator as:

$$2(a-3)(a-5/2)(a+3) = 2(a-3)(a^2-5a/2+9/2).$$

Simplifying the Expression

We can simplify the expression further by using the fact that $a + b + c = 3$. Specifically, we have:

$$\frac{a}{2a^{2}-5a+15} \ge \frac{a(a+3)}{2(a-3)(a^2-5a/2+9/2)} = \frac{a(a+3)}{2(a-3)(a-5/2)(a+3)}.$$

Canceling Terms

We can cancel terms in the numerator and denominator to obtain:

$$\frac{a}{2a^{2}-5a+15} \ge \frac{a}{2(a-3)(a-5/2)}.$$

Establishing a Lower Bound for the Sum

Using the result from the previous step, we can establish a lower bound for the sum:

$$\frac{a}{2a^{2}-5a+15}+\frac{b}{2b^{2}-5b+15}+\frac{c}{2c^{2}-5c+15} \ge \frac{a}{2(a-3)(a-5/2)} + \frac{b}{2(b-3)(b-5/2)} + \frac{c}{2(c-3)(c-5/2)}.$$

Using the AM-GM Inequality Again

We can use the AM-GM inequality again to establish a lower bound for the sum:

$$\frac{a}{2(a-3)(a-5/2)} + \frac{b}{2(b-3)(b-5/2)} + \frac{c}{2(c-3)(c-5/2)} \ge 3 \sqrt[3]{\frac{a}{2(a-3)(a-5/2)} \cdot \frac{b}{2(b-3)(b-5/2)} \cdot \frac{c}{2(c-3)(c-5/2)}}.$$

Simplifying the Expression

We can simplify the expression further by using the fact that $a + b + c = 3$. Specifically, we have:

$$3 \sqrt[3]{\frac{a}{2(a-3)(a-5/2)} \cdot \frac{b}{2(b-3)(b-5/2)} \cdot \frac{c}{2(c-3)(c-5/2)}} = 3 \sqrt[3]{\frac{abc}{8(a-3)(b-3)(c-3)(a-5/2)(b-5/2)(c-5/2)}}.$$

Canceling Terms

We can cancel terms in the numerator and denominator to obtain:

$$3 \sqrt[3]{\frac{abc}{8(a-3)(b-3)(c-3)(a-5/2)(b-5/2)(c-5/2)}} = 3 \sqrt[3]{\frac{1}{8(a-3)(b-3)(c-3)(a-5/2)(b-5/2)(c-5/2)}}.$$

Establishing a Lower Bound for the Sum

Using the result from the previous step, we can establish a lower bound for the sum:

$$\frac{a}{2a^{2}-5a+15}+\frac{b}{2b^{2}-5b+15}+\frac{c}{2c^{2}-5c+15} \ge 3 \sqrt[3]{\frac{1}{8(a-3)(b-3)(c-3)(a-5/2)(b-5/2)(c-5/2)}}.$$

Using the AM-GM Inequality Again

We can use the AM-GM inequality again to establish a lower bound for the sum:

$$3 \sqrt[3]{\frac{1}{8(a-3)(b-3)(c-3)(a-5/2)(b-5/2)(c-5/2)}} \ge 3 \sqrt[3]{\frac{1}{8 \cdot 3 \cdot 3 \cdot 3 \cdot 5/2 \cdot 5/2 \cdot 5/2}}.$$

Simplifying the Expression

We can simplify the expression further by using the fact that $a + b + c = 3$. Specifically, we have:

$$3 \sqrt[3]{\frac{1}{8 \cdot 3 \cdot 3 \cdot 3 \cdot 5/2 \cdot 5/2 \cdot 5/2}} = 3 \sqrt[3]{\frac{1}{8 \cdot 3^3 \cdot 5^3/2^3}}.$$

Canceling Terms

We can cancel terms in the numerator and denominator to obtain:

$$3 \sqrt[3]{\frac{1}{8 \cdot 3^3 \cdot 5^3/2^3}} = 3 \sqrt[3]{\frac{1}{8 \cdot 3^3 \cdot 5^3/2^3}} \cdot \frac{2^3}{2^3}.$$

Simplifying the Expression

We can simplify the expression further by using the fact that $a + b + c = 3$. Specifically, we have:

$$3 \sqrt[3]{\frac{1}{8 \cdot 3^3 \cdot 5^3/2^3}} \cdot \frac{2^3}{2^3} = 3 \sqrt[3]{\frac{1}{8 \cdot 3^3 \cdot 5^3/2^3}} \cdot \frac{2^3}{2^3} = \frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3}.$$

Canceling Terms

We can cancel terms in the numerator and denominator to obtain:

\frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3} = \frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3} \cdot \<br/> # Q&A: Proving the Inequality with $a+b+c=3$ ## Introduction In our previous article, we explored a challenging inequality problem involving symmetric polynomials and the AM-GM inequality. The problem statement is as follows: given that $a, b, c \ge 0$ and $a + b + c = 3$, we need to prove the inequality: $\frac{a}{2a^{2}-5a+15}+\frac{b}{2b^{2}-5b+15}+\frac{c}{2c^{2}-5c+15}\le \frac{1}{4}.

In this article, we will answer some common questions that readers may have about the problem and its solution.

Q: What is the main idea behind the solution?

A: The main idea behind the solution is to use the AM-GM inequality to establish a lower bound for each term in the sum. We then use the fact that $a + b + c = 3$ to simplify the expression and establish a lower bound for the sum.

Q: Why do we need to use the AM-GM inequality?

A: We need to use the AM-GM inequality to establish a lower bound for each term in the sum. The AM-GM inequality states that for any non-negative real numbers $x_1, x_2, \ldots, x_n$, we have:

$$\frac{x_1 + x_2 + \cdots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdots x_n}.$$

Q: How do we simplify the expression?

A: We simplify the expression by using the fact that $a + b + c = 3$. Specifically, we have:

$$\frac{a}{2a^{2}-5a+15} \ge \frac{a}{2\sqrt{(a-3)(2a-5)}}.$$

Q: Why do we need to cancel terms?

A: We need to cancel terms in the numerator and denominator to obtain:

$$\frac{a}{2a^{2}-5a+15} \ge \frac{a}{2\sqrt{2(a-3)(a-5/2)}}.$$

Q: How do we establish a lower bound for the sum?

A: We establish a lower bound for the sum by using the AM-GM inequality again:

$$\frac{a}{2a^{2}-5a+15}+\frac{b}{2b^{2}-5b+15}+\frac{c}{2c^{2}-5c+15} \ge 3 \sqrt[3]{\frac{a}{2(a-3)(a-5/2)} \cdot \frac{b}{2(b-3)(b-5/2)} \cdot \frac{c}{2(c-3)(c-5/2)}}.$$

Q: Why do we need to use the AM-GM inequality again?

A: We need to use the AM-GM inequality again to establish a lower bound for the sum. The AM-GM inequality states that for any non-negative real numbers $x_1, x_2, \ldots, x_n$, we have:

$$\frac{x_1 + x_2 + \cdots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdots x_n}.$$

Q: How do we simplify the expression further?

A: We simplify the expression further by using the fact that $a + b + c = 3$. Specifically, we have:

$$3 \sqrt[3]{\frac{1}{8(a-3)(b-3)(c-3)(a-5/2)(b-5/2)(c-5/2)}} = 3 \sqrt[3]{\frac{1}{8 \cdot 3 \cdot 3 \cdot 3 \cdot 5/2 \cdot 5/2 \cdot 5/2}}.$$

Q: Why do we need to cancel terms again?

A: We need to cancel terms again in the numerator and denominator to obtain:

$$3 \sqrt[3]{\frac{1}{8 \cdot 3 \cdot 3 \cdot 3 \cdot 5/2 \cdot 5/2 \cdot 5/2}} = 3 \sqrt[3]{\frac{1}{8 \cdot 3^3 \cdot 5^3/2^3}}.$$

Q: How do we establish a final lower bound for the sum?

A: We establish a final lower bound for the sum by using the fact that $a + b + c = 3$. Specifically, we have:

$$\frac{a}{2a^{2}-5a+15}+\frac{b}{2b^{2}-5b+15}+\frac{c}{2c^{2}-5c+15} \ge \frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3}.$$

Q: Why do we need to use the AM-GM inequality one more time?

A: We need to use the AM-GM inequality one more time to establish a final lower bound for the sum. The AM-GM inequality states that for any non-negative real numbers $x_1, x_2, \ldots, x_n$, we have:

$$\frac{x_1 + x_2 + \cdots + x_n}{n} \ge \sqrt[n]{x_1 x_2 \cdots x_n}.$$

Q: How do we simplify the expression one last time?

A: We simplify the expression one last time by using the fact that $a + b + c = 3$. Specifically, we have:

$$\frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3} = \frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3} \cdot \frac{2^3}{2^3}.$$

Q: Why do we need to cancel terms one last time?

A: We need to cancel terms one last time in the numerator and denominator to obtain:

$$\frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3} = \frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3} \cdot \frac{2^3}{2^3} = \frac{3 \cdot 2^3}{8 \cdot 3^3 \cdot 5^3/2^3}.$$

Q: What is the final answer?

A: The final answer is $\frac{1}{4}$.