×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Q3....................................................(just to reach min. char. limit


4 years ago

mycroft holmes
272 Points
							$\because a\ge -1, (a+1)(a^2-2a+1)\ge 0$ and hence we have $a^3 -a^2 -a +1 \ge 0 \Rightarrow a^3+1 \ge a^2+a$ and so $\sum a^2+a \le \sum a^3+1 = 1+3 = 4$ i.e. $a+b+c+a^2+b^2+c^2 \le 4$ Equality can occur when $a,b,c \in \{1,-1 \}$ and since require that $a^3+b^3+c^3 = 1$ this means we can choose a=b=1 and c=-1 and permutations of this as the equality case.

4 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions