Let $a,b,$ and $c$ be the lengths of the sides of a triangle, prove that $$a(b^2+c^2-a^2)+b(a^2+c^2-b^2)+c(a^2+b^2-c^2) \le 3abc.$$

I can't really factor this into something nice. Also using AM-GM or Cauchy-Schwarz doesn't look like it will help. I am thinking we need to bound the left and right side with something, but I don't know how to.

  • 3
    $\begingroup$ Hint: try the cosine rule $\endgroup$ – Nikunj Jan 21 '16 at 13:33

Hint: Expand the following, divide by 2, and rearrange: $$ (a+b-c)(a-b)^2+(b+c-a)(b-c)^2+(c+a-b)(c-a)^2\geq 0. $$

How I thought of this: I was thinking of a multinomial with degree $3$ that is symmetrical in $a,b,c$, is nonnegative, uses the triangle inequality, and goes away when $a=b=c$. The above was the first guess.


Note: Using the cosine rule, this inequality can be written as:


Now, we have to prove that $cosA+cosB+cosC\le3/2$

Proof: $f(A,B) = cos(A) + cos(B) + cos(π-A-B)$

$f(A,B) = cos(A) + cos(B) - cos(A+B)$

within the region R:



Since it's an open region, we know the maximum cannot occur anywhere along the boundary of R and must occur at some critical point(s) in the interior.

So we just have to find the point where the total derivative is 0. Of course, A and B aren't dependent on each other, so we can just set the two partials equal to zero to find the critical point(s):

$$sin(A+B) - sin(A) = 0$$

$$sin(A+B) - sinB = 0$$

So $sinA = sinB$, thus A=B

But $sinA = sin2A = 2sinAcosA$

Thus $$cosA = 1/2$$

Note: we can divide by sinA because A≠0. This also excludes A=0 as a solution, since A=0 is in the boundary of R, which is not a part of our valid region.

$$A = π/3$$

So the maximum is at $$A=B=C=π/3$$.

Plugging that in, we know the maximum value of f is:

$3cos(π/3) = 3/2$

Thus for angles A,B,C of a triangle:

$$cos(A)+cos(B)+cos(C) = f(A,B) = ≤ 3/2$$ .


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