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Xn ) if and only if (b) ≺ (a). Equality takes place only when (b) and (a) are identical or when all the xi s are equal. Before going through the proof, which is quite diﬃcult, let us look at some examples. 1 is not satisﬁed, but we can see that [2, 0, 0] ≥ [1, 1, 0], which is equivalent to x2 + y 2 + z 2 ≥ xy + yz + zx. In the same way, we can see that 1. x2 + y 2 ≥ 2xy ⇔ [2, 0] ≥ [1, 1], 2. x3 + y 3 + z 3 ≥ 3xyz ⇔ [3, 0, 0] ≥ [1, 1, 1], 3. x5 + y 5 ≥ x3 y 2 + x2 y 3 ⇔ [5, 0] ≥ [3, 2], 4. 8 Muirhead’s theorem 45 and all these inequalities are satisﬁed if we take for granted Muirhead’s theorem.

Now, we will consider some criteria to decide whether a function is convex. 5. 3 Proof. Suppose that f is convex and let A = (x1 , y1 ) and B = (x2 , y2 ) be two points in the set U = {(x, y) | a ≤ x ≤ b, f (x) ≤ y}. To prove that tB + (1 − t)A = (tx2 + (1 − t)x1 , ty2 + (1 − t)y1 ) belongs to U , it is suﬃcient to demonstrate that a ≤ tx2 + (1 − t)x1 ≤ b and f (tx2 + (1 − t)x1 ) ≤ ty2 + (1 − t)y1 . The ﬁrst condition follows immediately since x1 and x2 belong to [a, b]. As for the second condition, since f is convex, it follows that f (tx2 + (1 − t)x1 ) ≤ tf (x2 ) + (1 − t)f (x1 ).

Hence, if in the triangle ABC we have ∠A > ∠B, then BC > CA. 1. (i) If a, b, c are positive numbers with a < b + c, b < c + a and c < a + b, then a triangle exists with side lengths a, b and c. (ii) To be able to construct a triangle with side lengths a ≤ b ≤ c, it is suﬃcient that c < a + b. (iii) It is possible to construct a triangle with sides of length a, b and c if and only if there are positive numbers x, y, z such that a = x + y, b = y + z and c = z + x. 2. (i) If it is possible to construct a triangle with side-lengths b< √a < √ √ c, then it is possible to construct a triangle with side-lengths a < b < c.