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

Revision as of 04:50, 17 June 2021

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 $m \geq 4$ .

$Kris17$

Solution 2

By Trivial inequality, $(x - y)^2 \geq 0 \iff 2(x^2 - xy + y^2) \geq 0 \iff \frac{2}{x^2 + y^2} \geq \frac{1}{x^2 - xy + y^2}.$

Then by multiplying by $x + y$ on both sides, we use the Trivial Inequality again to obtain $2(x^2 + y^2) \geq (x + y)^2$ which means $\frac{x+y}{x^2 - xy + y^2} \leq \frac{2(x + y)}{x^2 + y^2} \leq \frac{2\sqrt{2(x^2+y^2)}}{x^2 + y^2}$ which after simplifying, proves the problem.

@@ Line 22: / Line 22: @@
 <math>Kris17</math>
+==Solution 2==
+By Trivial inequality,
+<cmath> (x - y)^2 \geq 0 \iff 2(x^2 - xy + y^2) \geq 0 \iff \frac{2}{x^2 + y^2} \geq \frac{1}{x^2 - xy + y^2}.</cmath>
+Then by multiplying by <math>x + y</math> on both sides, we use the Trivial Inequality again to obtain <math>2(x^2 + y^2) \geq (x + y)^2</math> which means <cmath>\frac{x+y}{x^2 - xy + y^2} \leq \frac{2(x + y)}{x^2 + y^2} \leq \frac{2\sqrt{2(x^2+y^2)}}{x^2 + y^2}</cmath> which after simplifying, proves the problem.

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

Revision as of 04:50, 17 June 2021

Problem

Solution

Solution 2