Question
Assertion: If $a, b, c$ are distinct positive real numbers and $a^{2}+b^{2}+c^{2}=1$, then $a b+b c+c a$ is less than 1 . Reason: A.M. >G.M. for unequal numbers
Step 1
This is known as the AM-GM inequality. Show more…
Show all steps
Your feedback will help us improve your experience
Ankit Singh and 80 other Calculus 2 / BC 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
If $a, b, c$ are distinct positive real numbers and $a^{2}+b^{2}$ $+c^{2}=1$, then $a b+b c+c a$ is (A) less than 1 (B) equal to 1 (C) greater than 1 (D) any real number
Statement-1: If $a, b, c$ are three positive real numbers such that $a+c \neq b$ and $$ \frac{1}{a}+\frac{1}{a-b}+\frac{1}{c}+\frac{1}{c-b}=0 $$ then $a, b, c$ are in $H, P$. Statement-2: If $a, b, c$ are distinct positive real numbers such that $$ a(b-c) x^{2}+b(c-a) x y+c(a-b) y^{2} $$ is a perfect square, then $a, b, c$ are in $H$.P.
If $a, b$ and $c$ are positive and unequal, show that $(a+b+c)^{2}<3\left(a^{2}+b^{2}+c^{2}\right)$.
Algebraic Functions, Equations and Inequalities
Other equations and inequalities
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD