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

(Created page with "==Problem== Prove that the inequality <cmath> \frac{ x+y}{x^2-xy+y^2 } \leq \frac{ 2\sqrt 2 }{\sqrt{ x^2 +y^2 } } </cmath> holds for all real numbers <math>x</math> and <math...")
 
Line 16: Line 16:
 
Now squaring both sides of the inequality, we get:
 
Now squaring both sides of the inequality, we get:
 
<cmath> \frac{m}{(m-3)^2 } \leq \frac{8}{m-2} </cmath>
 
<cmath> \frac{m}{(m-3)^2 } \leq \frac{8}{m-2} </cmath>
−
after cross multiplication and some simplification we get:
+
after cross multiplication and simplification we get:
 
<math>7m^2 -46m + 72 \geq 0</math>
 
<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>

Revision as of 23:14, 16 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, $7(m-4)^2 +10(m-4) \geq 0$ The above is always true since $m \geq 4$.


$Kris17$

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