My solution

American Mathematical Monthly Problem 11435 by Panagiote Ligouras, Noci, Italy.

Given a triangle ABC with sidelengths a, b, c, inradius r and circumradius R.

Prove the inequality a²bc

(a+b) (a+c) + b²ca

(b+c) (b+a)+ c²ab

(c+a) (c+b) ≤ 9 2rR.

Solution by Darij Grinberg.

Let ∆ be the area of triangle ABC, and s = 1

2(a+b+c) its semiperimeter. It is known that ∆ = sr,so that r= ∆

s = ∆

1

2(a+b+c)

= 2∆

a+b+c, and it is also known that R = abc

4∆. Thus, 9

2rR = 9

2 · 2∆

a+b+c · abc

4∆ = 9abc

4 (a+b+c). Thus, the inequality that we have to prove,

a²bc

(a+b) (a+c) + b²ca

(b+c) (b+a)+ c²ab

(c+a) (c+b) ≤ 9 2rR, rewrites as

a²bc

(a+b) (a+c) + b²ca

(b+c) (b+a) + c²ab

(c+a) (c+b) ≤ 9abc 4 (a+b+c). Upon multiplication by a+b+c

abc , this becomes a(a+b+c)

(a+b) (a+c) + b(a+b+c)

(b+c) (b+a)+ c(a+b+c) (c+a) (c+b) ≤ 9

4. (1)

Now, consider a new triangle with sidelengthsa⁰ =b+c, b⁰ =c+a, c⁰ =a+b (such a triangle actually exists because b+c, c+a, a+b satisfy the triangle inequalities: for instance, (c+a) + (a+b)>(b+c)). LetA⁰, B⁰, C⁰ be the three angles of this triangle, and lets⁰ = 1

2(a⁰+b⁰+c⁰) be the semiperimeter of this triangle. Then, s⁰ = 1

2(a⁰+b⁰+c⁰) = 1

2((b+c) + (c+a) + (a+b)) =a+b+c, and a standard formula yields

cos² A⁰

2 = s⁰(s⁰−a⁰)

b⁰c⁰ = (a+b+c) ((a+b+c)−(b+c))

(c+a) (a+b) = a(a+b+c) (a+b) (a+c), and similarly cos² B⁰

2 = b(a+b+c)

(b+c) (b+a) and cos² C⁰

2 = c(a+b+c)

(c+a) (c+b). Hence, the inequality in question, (1), is equivalent to

cos² A⁰

2 + cos² B⁰

2 + cos²C⁰ 2 ≤ 9

4. 1

But this is easy using Jensen’s inequality: The function [0, π]→R, ϕ 7→cos² ϕ

2, is concave (since cos² ϕ

2 = cosϕ+ 1

2 , and the function [0, π]→R, ϕ7→cosϕ is concave), and thus the Jensen inequality yields

cos² A⁰

2 + cos² B⁰

2 + cos² C⁰

2 ≤3 cos²

A⁰+B⁰+C⁰ 3

2 = 3 cos² A⁰+B⁰+C⁰ 6

= 3 cos²π

6 (since A⁰,B⁰,C⁰ are angles of a triangle)

= 3· 3 4 = 9

4, qed.

2