inequality

**earthboy** · January 26th 2013, 10:35 PM

Originally Posted by mulfer

$Click image for larger version. Name: inequality1.jpg Views: 4 Size: 33.0 KB ID: 26709$

Small advice @ mulfer

lease learn LATEX

We have to prove:
$a^2+b^2+c^2 \geq a+b+c$
It is equivalent to proving:
$a^2+b^2+c^2-a-b-c \geq 0$
$\implies a(a-1)+b(b-1)c(c-1) \geq 0$ So we are done if we are able to prove this.
now,
if $x$ is a positive real number,
if $x>1$ then $x(x-1)>(x-1)$ (obvious)
if $x=1$ then $x(x-1)=(x-1)$ (obvious)
if $0<x<1$ then $(x-1)<0$ so, $x<1$ $\implies x(x-1)>x-1$ (multiplying both sides by a negative number i.e $(x-1)$ changes the sign)
so for all case, $x(x-1)\geq x-1$
similarly,
$a(a-1) \geq a-1$
$b(b-1) \geq b-1$
$c(c-1) \geq c-1$
summing these we get:
$a(a-1)+b(b-1)c(c-1) \geq (a-1)+(b-1)+(c-1) = (a+b+c)-3$
so we will be done if we prove $(a+b+c)-3 \geq 0$ or $a+b+c \geq 3$
by the AM-GM inequality,
$(a+b+c)\geq 3(abc)^{1/3}=3$ # as $abc=1$
proved

**mulfer** · January 27th 2013, 04:14 AM

thanks for replying. I'd be glad to learn the latex soon.
please, check my solution:
$inequality-.jpg$

**earthboy** · January 27th 2013, 05:32 AM

Originally Posted by mulfer

thanks for replying. I'd be glad to learn the latex soon.
please, check my solution:
$Click image for larger version. Name: متباينة.jpg Views: 11 Size: 71.3 KB ID: 26722$

Yup..Its seems alright to me

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