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

Latest revision as of 08:38, 14 March 2023

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

Again, since the inequality is homogenous, we can assume WLOG that $x^2+y^2=8$ .

By AM-GM we gave $xy\leq4=\frac{x^2+y^2}{2}$ and by QM-AM we have that $x+y\leq4=2\sqrt{\frac{x^2+y^2}{2}}$ .

Substituting we have $\frac{x+y}{x^2-xy+y^2}\leq\frac{4}{4}=\frac{2\sqrt{2}}{\sqrt{8}}=\frac{2\sqrt{2}}{\sqrt{x^2+y^2}}$

$DVDTSB$

Solution 3

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

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 24: / Line 24: @@
 ==Solution 2==
+Again, since the inequality is homogenous, we can assume WLOG that <math>x^2+y^2=8</math>.
+By AM-GM we gave <math>xy\leq4=\frac{x^2+y^2}{2}</math> and by QM-AM  we have that <math>x+y\leq4=2\sqrt{\frac{x^2+y^2}{2}}</math>.
+Substituting we have
+<cmath>\frac{x+y}{x^2-xy+y^2}\leq\frac{4}{4}=\frac{2\sqrt{2}}{\sqrt{8}}=\frac{2\sqrt{2}}{\sqrt{x^2+y^2}}</cmath>
+<math>DVDTSB</math>
+==Solution 3==
 By Trivial Inequality,

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