Alzebra : 1023

Problem 

The positive reals x, y satisfy x² + y³ ≥ x³ + y⁴.   Show that x³ + y³ ≤ 2.

 

Solution

If x, y both ≤ 1, then the inequality is obvious. If x, y both > 1, then they do not satisfy the condition. 

So either x ≥ 1 ≥ y or y ≥ 1 ≥ x. 

We show first that x³ + y³ ≤ x² + y². 

In the first case we have x² + y²(1-y) ≥ x² + y³(1-y) ≥ x³ + y⁴ - y⁴ = x³, 

so x² + y² ≥ x³ + y³, as required. 

In the second case we have x² ≥ x³ + y³(y-1) ≥ x³ + y²(y-1), so again x² + y² ≥ x³ + y³.

Now by Cauchy-Scwartz, x² + y² = x^3/2x^1/2 + y^3/2y^1/2 ≤ √(x³+y³) √(x+y). 

But x² + y² ≥ x³ + y³, so x² + y² ≤ x + y. 

But 2xy ≤ (x²+y²), so (x + y)² ≤ & 2(x² + y²) and 

we have just shown that 2(x² + y²) ≤ 2(x + y), so x + y ≤ 2. 

Thus we have x³ + y³ ≤ x² + y² ≤ x + y ≤ 2.