Ono's inequality

In mathematics, Ono's inequality is a theorem about triangles in the Euclidean plane. In its original form, as conjectured by Takashi Ono in 1914, the inequality is actually false; however, the statement is true for acute triangles, as shown by F. Balitrand in 1916.

Statement of the inequality
Consider an acute triangle (meaning a triangle with three acute angles) in the Euclidean plane with side lengths a, b and c and area S. Then


 * $$27 (b^2 + c^2 - a^2)^2 (c^2 + a^2 - b^2)^2 (a^2 + b^2 - c^2)^2 \leq (4 S)^6.$$

This inequality fails for general triangles (to which Ono's original conjecture applied), as shown by the counterexample $$a=2, \, \, b=3, \, \, c=4, \, \, S=3\sqrt{15}/4.$$

The inequality holds with equality in the case of an equilateral triangle, in which up to similarity we have sides $$1,1,1$$ and area $$\sqrt{3}/4.$$

Proof
Dividing both sides of the inequality by $$64(abc)^4$$, we obtain:

$$27 \frac{(b^2 + c^2 - a^2)^2}{4b^2c^2} \frac{(c^2 + a^2 - b^2)^2}{4a^2c^2} \frac{(a^2 + b^2 - c^2)^2}{4a^2b^2} \leq \frac{4S^2}{b^2c^2} \frac{4S^2}{a^2c^2} \frac{4S^2}{a^2b^2}$$

Using the formula $$S= \tfrac12 bc\sin{A}$$ for the area of triangle, and applying the cosines law to the left side, we get:
 * $$27 (\cos{A} \cos{B} \cos{C})^2 \leq (\sin{A} \sin{B} \sin{C})^2$$

And then using the identity $$\tan{A} + \tan{B} + \tan{C} = \tan{A} \tan{B} \tan{C}$$ which is true for all triangles in euclidean plane, we transform the inequality above into:
 * $$27 (\tan{A} \tan{B} \tan{C}) \leq (\tan{A} + \tan{B} + \tan{C})^3$$

Since the angles of the triangle are acute, the tangent of each corner is positive, which means that the inequality above is correct by AM-GM inequality.