Maths Olympiad Prep

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Problem 700

AMC 10/12, early questions
Algebra Difficulty 3.2 Prove it Junior Turkish Mathematical Olympiad · Turkey

Prove that
1≤(x+y)(x3+y3)(x2+y2)2≤98 1 \le \frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2} \le \frac{9}{8}
for all positive real numbers xx and yy.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

The Cauchy-Schwarz inequality implies that (x+y)(x3+y3)≥(x2+y2)2(x+y)(x^3+y^3) \ge (x^2+y^2)^2. Therefore,
1≤(x+y)(x3+y3)(x2+y2)2. 1 \le \frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2}.
As 0≤((x−y)2−2xy)20 \le ((x-y)^2 - 2xy)^2, we have
4xy(x−y)2≤(x−y)4+4x2y2. 4xy(x-y)^2 \le (x-y)^4 + 4x^2y^2.
⇒8x3y+8y3x≤x4+18x2y2+y4.⇒8x3y+8y3x+8x4+8y4≤9x4+18x2y2+9y4. \Rightarrow 8x^3y + 8y^3x \le x^4 + 18x^2y^2 + y^4. \Rightarrow 8x^3y + 8y^3x + 8x^4 + 8y^4 \le 9x^4 + 18x^2y^2 + 9y^4.
⇒8(x+y)(x3+y3)≤9(x2+y2)2. Thus, \Rightarrow 8(x + y)(x^3 + y^3) \le 9(x^2 + y^2)^2. \text{ Thus,}
(x+y)(x3+y3)(x2+y2)2≤98. \frac{(x+y)(x^3+y^3)}{(x^2+y^2)^2} \le \frac{9}{8}.

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