Question

If a,b,c are real numbers such that:

\(a^2 + b^2 + c^2 = 1\)

prove the inequality:

\(\cfrac{-1}{2} \le ab+bc+ca \le 1\)

Answer 1

@Miracrown @ganeshie8

Answer 2

Okay thanks @Miracrown I think I got that method also. Thanks for your help friends ..

Answer 3

You're welcome.

Answer 4

@vishweshshrimali5 - I got the first part of the inequality :

\((a+b+c)^2 \ge 0 \\
1 + 2(ab + bc + ca) \ge 0 \quad ; \boxed{(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) } \\
2(ab + bc + ca) \ge - 1 \quad ;\space \\

ab + bc + ca \ge \cfrac{-1}{2} \\ \)

Answer 5

Did you use AM-GM inequality to solve for the second inequality?

Answer 6

No

Answer 7

Hmm, then ?

Answer 8

We used this:

\((a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0\)

Answer 9

\(a^2 + b^2 - 2ab + b^2 + c^2 -2bc + c^2 + a^2 -2ca \ge 0\)
\(2(a^2 + b^2 + c^2) -2(ab +bc+ca) \ge 0\)
\(ab + bc + ca \le 1\)

Cool! Got it. Thanks.

Answer 10

No problem.

Answer 11

Though, I have another method for this.

Answer 12

AM GM Inequality will also work here. Though, what you have used here, is also correct, but just wanted to put this up here :

\[\cfrac{a^2 + b^2}{2} \ge \sqrt{a^2b^2} \\
\boxed{\color{Blue}{\cfrac{a^2 + b^2}{2} \ge ab}} ---------(1) \]
\[\cfrac{b^2 + c^2}{2} \ge \sqrt{b^2 c^2} \\
\boxed{\color{blue}{\cfrac{b^2 + c^2}{2} \ge bc }} ---------(2) \]

\[\cfrac{c^2 + a^2}{2} \ge \sqrt{c^2 a^2} \\
\boxed{\color{blue}{\cfrac{c^2 + a^2}{2} \ge ca }} ---------(3) \]

Adding equations 1 , 2 and 3 :

\[\cfrac{2(a^2 + b^2 + c^2)}{2} \ge ab + bc + ca \\
\cfrac{\cancel{2} (1)}{\cancel{2}} \ge ab + bc + ca \quad ; \boxed{a^2 + b^2 + c^2 = 1 } \space \text{(Given)} \\
1 \ge ab + bc + ca \\
\text{OR} \quad \boxed{\color{red}{\bf{-\cfrac{1}{2} \ge ab + bc + ca\le 1}}} \]

Answer 13

Great work @mathslover

Answer 14

Thank you! :-)