Math Olympiad Question?

Question

Suppsoe that a, b, c are sides of of a triangle. Prove that:
a^2(b + c - a) + b^2(c + a - b) + c^2(a + b - c) <= 3abc

I can't seem to do this! I've tried using algebra and triangle inequality rules. Should I use trignometry?!?!

Please provide me with work and an answer.

Thanks! Merry Christmas.

FOR THE MOMENT, UNTIL I SAY "OKAY", PLEASE DO NOT ANSWER. I'm currently getting on to something, so I wish to do this on my own =D!

Thanks!

OKAY!

Here's my work. I'm not sure if it's "proved" though.

c^2 = a^2 + b^2 - (2ab cos C)

a^2 (a+b-c) + a^2 (b+c-a) +
b^2 (a+b-c) + b^2(c+a-b) -
(2ab cos C)(a+b-c) <= 3abc

(2)(a^2)(b) + (2)(b^2)(a) -
(2ab)(cosC)(a+b-c) <= 3abc

2a + 2b - (2)(cos C)(a+b-c) <= 3c

2(a+b) - (2)(cos C)(a+b-c) <= 3c
(2/3)(a+b) - (2/3)(cos C)(a+b-c) <= c

a+b > c
(2/3)(a+b) > (2/3)c

Going back to the inequility I have now, I subtract (2/3)(a+b).

c - (# greater than (2/3)c) = # less than (1/3)c

Since c is the maximum, the inequalty says:

-(2)(cos C)(a+b-c) < (1/3)(c)

The left side expression is negative for all a, b, or c and the right side is positive, so the inequilty is proved to be true.

Now. At the end of my proof..... I subtracted (2/3)(a+b) from c. How doI know that c is always greater than (2/3)(a+b)? Is it even correct?

If my proof is wrong, please start from there, when I haven't subtracted (2/3)(a+b) yet.

If you HAVE to, do it your way.

Also, is it smart to do an inequilty within an inequilty like I did?

Answer 1

"How do I know that c is always greater than (2/3)(a+b)? Is it even correct?"

No, it is not correct. (It is easy to find an example when it is not valid.
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I have additional sub-question:

For the angles of any triangle always
cosA + cosB + cosC ≤ 1.5

Do you have a reference or proof of the above statement?

(That is a clue to solution of the problem.)

-

c² = a² + b² - 2ab cosC
c³ = a²c + b²c - 2abc cosC

a²c + b²c - c³ = 2abc cosC
ab² + ac² - a³ = 2abc cosA
bc² + a²b - b³ = 2abc cosB

a²c + b²c - c³ + ab² + ac² - a³ + bc² + a²b - b³ = 2abc (cosA + cosB + cosC)

cosA + cosB + cosC ≤ 1.5

a²c + b²c - c³ + ab² + ac² - a³ + bc² + a²b - b³ ≤ 3abc
(a²b + a²c - a³) + (b²c + ab² - b³) + (ac² + bc² - c³) ≤ 3abc
a²(b + c - a) + b²(c + a - b) + c²(a + b - c) ≤ 3abc
-

Answer 2

Once again the symmetry of the problem points to the case where a=b=c as the limiting case. And we verify that in this case, LHS = RHS.

So let's prove that when we deviate from a=b=c,
then LHS < RHS
Let's start with a=b=c = x
Now let's make a = x + d, while b = c = x

LHS = (x+d)^2 *(x-d) + x^2 (x+d) + x^2 (x + d)
= (x^2 + 2dx + d^2)*(x - d) + 2(x^3 + dx^2)
= 3x^3 + 3dx^2 -2d^2 x - d^3

RHS = 3x^2 (x+d)
= 3x^3 + 3dx^2

So RHS - LHS = 2d^2 x + d^3
= d^2 *(2x + d)

Clearly the difference is positive as long as d > -2x.
But to maintain positive sides of the triangle, d > -x,
so RHS - LHS > 0 for d ≠ 0

This is not comphrehensive proof, but I doubt very much that this is a test of complicated algebra and trig.