Question
Prove that if $a, b$ and $c$ are positive real numbers, then $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9$.
Step 1
Step 1: Recognize the inequality to be proven: We need to show that for positive real numbers \(a\), \(b\), and \(c\), the inequality \((a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9\) holds. Show more…
Show all steps
Your feedback will help us improve your experience
Shima Shaw and 97 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$$ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>9 $$
$$ (a+b+c)(b c+c a+a b)>9 a b c $$
$$ \frac{b c}{a}+\frac{c a}{b}+\frac{a b}{c} \geq a+b+c $$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD