Can Sum to Product Inequalities Hold for Non-Negative Reals?

In summary, sum to product inequalities are mathematical inequalities that involve the sum and product of variables. To solve them, the sum must be isolated and rewritten using trigonometric identities before using algebraic manipulations to find the variable. These inequalities are important in establishing relationships and solving for unknown values in various fields. They can also be applied to other types of equations by using trigonometric identities to rewrite them as sums or products.
  • #1
lfdahl
Gold Member
MHB
749
0
Given non-negative reals, $\alpha_i$, where $i = 1,2,...,n.$

Prove, that

$\alpha_1+\alpha_2+...+\alpha_n \leq \frac{1}{2}$ $\Rightarrow$ $(1-\alpha_1)(1-\alpha_2)...(1-\alpha_n) \geq \frac{1}{2}.$
 
Mathematics news on Phys.org
  • #2
Immediate consequence of Weiertrass inequality:

$$\prod_{1 \le k \le n}(1-a_k) \ge 1-\sum_{1 \le k \le n}a_k \ge 1-\frac{1}{2} = \frac{1}{2}.$$

There's an elegant proof of Weierstrass inequality.
 
  • #3
June29 said:
Immediate consequence of Weiertrass inequality:

$$\prod_{1 \le k \le n}(1-a_k) \ge 1-\sum_{1 \le k \le n}a_k \ge 1-\frac{1}{2} = \frac{1}{2}.$$

There's an elegant proof of Weierstrass inequality.
Thankyou for a clever solution, June29, and for your participation!(Cool)
 
  • #4
Can anyone prove the above statement by induction? (Wave)
 
  • #5
lfdahl said:
Can anyone prove the above statement by induction? (Wave)

It's obviously true for $n=1$ since we have $ {\alpha}_1 \leqslant \frac{1}{2} = 1-\frac{1}{2} \implies 1-\alpha_1 \geqslant \frac{1}{2}. $

Now, suppose it's true for $n = k \in \mathbb{N}$. We shall prove that it's true for $n=k+1$. $\displaystyle \begin{aligned} \frac{1}{2} & \leqslant 1- \sum_{1 \leqslant j \leqslant k+1}\alpha_j = 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j \leqslant 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j+a_{k+1}\sum_{1 \leqslant j \leqslant k} \alpha_j \\& =\left(1-\alpha_{k+1}\right)\left(1-\sum_{1 \leqslant j \leqslant k} \alpha_j\right)
\leqslant \left(1-\alpha_{k+1}\right) \prod_{1 \leqslant j \leqslant k} \left(1-\alpha_j \right) = \prod_{1 \leqslant j \leqslant k+1} \left(1-\alpha_j \right) \end{aligned} $

So it's true for $n=k+1$. Since it's true for $n=1, k+1$, it's true for all $n\in\mathbb{N}$.
 
Last edited by a moderator:
  • #6
June29 said:
It's obviously true for $n=1$ since we have $ {\alpha}_1 \leqslant \frac{1}{2} = 1-\frac{1}{2} \implies 1-\alpha_1 \geqslant \frac{1}{2}. $

Now, suppose it's true for $n = k \in \mathbb{N}$. We shall prove that it's true for $n=k+1$. $\displaystyle \begin{aligned} \frac{1}{2} & \leqslant 1- \sum_{1 \leqslant j \leqslant k+1}\alpha_j = 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j \leqslant 1-\alpha_{k+1}-\sum_{1 \leqslant j \leqslant k}\alpha_j+a_{k+1}\sum_{1 \leqslant j \leqslant k} \alpha_j \\& =\left(1-\alpha_{k+1}\right)\left(1-\sum_{1 \leqslant j \leqslant k} \alpha_j\right)
\leqslant \left(1-\alpha_{k+1}\right) \prod_{1 \leqslant j \leqslant k} \left(1-\alpha_j \right) = \prod_{1 \leqslant j \leqslant k+1} \left(1-\alpha_j \right) \end{aligned} $

So it's true for $n=k+1$. Since it's true for $n=1, k+1$, it's true for all $n\in\mathbb{N}$.

A nice solution, June29! Thankyou for your participation!

Please remember to hide your solution in SP tags. Other forum users might try to solve the challenge preferably without knowing your solution. Thankyou in advance!
 

What are sum to product inequalities?

Sum to product inequalities are mathematical inequalities involving the sum and product of two or more variables. These inequalities are used to establish relationships between variables and are commonly found in algebraic equations.

How do you solve sum to product inequalities?

To solve a sum to product inequality, you first need to isolate the sum on one side of the equation. Then, you can use trigonometric identities to rewrite the sum as a product. Finally, you can use algebraic manipulations to solve for the variable.

What are some common trigonometric identities used in sum to product inequalities?

Some common trigonometric identities used in sum to product inequalities include the double angle identities, half angle identities, and power reducing identities.

Why are sum to product inequalities important?

Sum to product inequalities are important because they allow us to establish relationships between variables and solve for unknown values. They are commonly used in real-world applications such as engineering, physics, and economics.

Can sum to product inequalities be applied to other types of equations?

Yes, sum to product inequalities can be applied to other types of equations such as exponential and logarithmic equations. By using trigonometric identities, these equations can be rewritten as sums or products, making them easier to solve.

Similar threads

Replies
5
Views
958
  • Differential Geometry
Replies
2
Views
538
  • General Math
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
982
  • Advanced Physics Homework Help
Replies
1
Views
684
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
2K
Replies
13
Views
1K
  • Precalculus Mathematics Homework Help
Replies
14
Views
2K
Back
Top