# prove that a^2 + b^2 + c^2 + 2abc <2

• Jul 23rd 2009, 12:36 AM
nh149
prove that a^2 + b^2 + c^2 + 2abc <2
for a,b,c real positive numbers such that a+b+c=2, a<b+c, b< c+a , c< a+b; show that

a^2 + b^2 + c^2 + 2abc < 2 .
first, i try to find abc and i got
abc< abc+a(c^2)+(a^2)b+c(b^2)+(c^2)b+ a(b^2)+abc
then i find (a^2)+(b^2)+(c^2)< 2(a^2)+2(b^2)+2(c^2)+2(ab+ac+bc)

i add both 2abc+(a^2)+(b^2)+(c^2) but i didnt end up with proved..
what can i do?(Wondering)
• Jul 23rd 2009, 11:28 PM
Opalg
Quote:

Originally Posted by nh149
for a,b,c real positive numbers such that a+b+c=2, a<b+c, b< c+a , c< a+b; show that

a^2 + b^2 + c^2 + 2abc < 2 .

We want to prove

$(1)\qquad a^2 + b^2 + c^2 + 2abc < 2$.

Multiply both sides by 4:

$(2)\qquad 4(a^2 + b^2 + c^2) + 8abc < 8$.

Since $a+b+c=2$, we can make (2) homogenous of degree three by writing it as

$(3)\qquad 2(a+b+c)(a^2 + b^2 + c^2) + 8abc < (a+b+c)^3$.

Multiply out the brackets in (3) and rearrange it a bit, and you will get

$(4)\qquad a^3+b^3+c^3 +2abc .

So far, the argument is reversible, so if we can prove (4) then (1) will also hold.

We are told that $a, $b< c+a$ and $c< a+b$. Therefore

$(5)\qquad (b+c-a)(c+a-b)(a+b-c) > 0$.

Multiply out the brackets in (5), rearrange it a bit, and you get exactly the inequality (4), as wanted.
• Jul 24th 2009, 01:11 AM
TheAbstractionist
Nice proof, Opalg. (Clapping)

Here is another proof (which I took all night thinking about).

$a\ <\ b+c$
$\implies\ a\ <\ a+b+c-a\ =\ 2-a$

$\implies\ 1-a\ >\ 0$

Similarly $1-b>0$ and $1-c>0.$

$\therefore\ (1-a)(1-b)(1-c)\ >\ 0$

$\implies\ 1-(a+b+c)+ab+bc+ca-abc\ >\ 0$

$\implies\ ab+bc+ca-abc\ >\ 1$

Now multiply through by 2, replace $2(ab+bc+ca)$ by $(a+b+c)^2-(a^2+b^2+c^2)$ and rearrange to get the desired inequality. (Smile)
• Jul 24th 2009, 02:28 AM
Opalg
Quote:

Originally Posted by TheAbstractionist
Nice proof, Opalg. (Clapping)

Here is another proof (which I took all night thinking about).

It kept me awake for quite a while too! (Sleepy)