Difference between revisions of "2004 JBMO Problems/Problem 1"
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after cross multiplication and 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> which is always true since <math>m \geq 4</math>. |
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+ | <math>Kris17</math> | ||
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+ | ==Solution 2== | ||
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+ | By Trivial Inequality, | ||
+ | <cmath> (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}.</cmath> | ||
− | <math> | + | 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. |
Revision as of 05:51, 17 June 2021
Problem
Prove that the inequality holds for all real numbers and , not both equal to 0.
Solution
Since the inequality is homogeneous, we can assume WLOG that xy = 1.
Now, substituting , we have:
, thus we have
Now squaring both sides of the inequality, we get:
after cross multiplication and simplification we get:
or, which is always true since .
Solution 2
By Trivial Inequality,
Then by multiplying by on both sides, we use the Trivial Inequality again to obtain which means which after simplifying, proves the problem.