# Inequality

If $$a$$ and $$b$$ are real numbers satisfying $$3a+4b=15$$, find the minimum value of $$\sqrt{a^2 + b^2}$$.

Note by Raj Mantri
1 year, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

For any real number a,b,x,y,we have inequality $$ax+by\leq \sqrt{a^2+b^2}\times\sqrt{x^2+y^2}$$ by using Cauchy-Swartz inequality.Hence, $$15=3a+4b\leq\sqrt{3^2+4^2}\times\sqrt{a^2+b^2}=5\sqrt{a^2+b^2}.$$
Therefore,the minimum value of $$\sqrt{a^2+b^2}$$ is $$\boxed{3}.$$

- 1 year, 8 months ago

using cauchy-swarch inequality in 3a+4b we get what ayush rai got u can check what is cauchy swarck inequality if u dont know

- 1 year, 8 months ago

So you were checking out my profile.How was my solution for the geometry problem.upvote if u like.

- 1 year, 8 months ago