Difference between revisions of "2004 JBMO Problems/Problem 1"

Revision as of 00:15, 17 December 2018

Problem

Prove that the inequality $\frac{ x+y}{x^2-xy+y^2 } \leq \frac{ 2\sqrt 2 }{\sqrt{ x^2 +y^2 } }$ holds for all real numbers $x$ and $y$ , not both equal to 0.

Solution

Since the inequality is homogeneous, we can assume WLOG that xy = 1.

Now, substituting $m = (x+y)^2$ , we have:

$m = x^2 + y^2 + 2xy = x^2 + y^2 + 2$ $\geq 2\sqrt {xy} + 2 = 4$ , thus we have $m \geq 4$

Now squaring both sides of the inequality, we get: $\frac{m}{(m-3)^2 } \leq \frac{8}{m-2}$ after cross multiplication and simplification we get: $7m^2 -46m + 72 \geq 0$ or, $7(m-4)^2 +10(m-4) \geq 0$ which is always true since $m \geq 4$ .

$Kris17$

@@ Line 18: / Line 18: @@
 after cross multiplication and simplification we get:
 <math>7m^2 -46m + 72 \geq 0</math>
-or, <math>7(m-4)^2 +10(m-4) \geq 0</math>
+or, <math>7(m-4)^2 +10(m-4) \geq 0</math> which is always true since <math>m \geq 4</math>.
-The above is always true since <math>m \geq 4</math>.
 <math>Kris17</math>

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Difference between revisions of "2004 JBMO Problems/Problem 1"

Revision as of 00:15, 17 December 2018

Problem

Solution