2. ## Re: inequality

Originally Posted by mulfer
Small advice @ mulferlease learn LATEX

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