Difference between revisions of "2008 Indonesia MO Problems/Problem 2"
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<cmath> \frac {1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} \ge \frac {2}{x + y + 2}.</cmath> | <cmath> \frac {1}{(1 + \sqrt {x})^{2}} + \frac {1}{(1 + \sqrt {y})^{2}} \ge \frac {2}{x + y + 2}.</cmath> | ||
− | ==Solution== | + | ==Solution 1== |
By the [[Cauchy-Schwarz Inequality]], <math>(1+1)(1+x) \ge (1 + \sqrt{x})^2</math> and <math>(1+1)(1+y) \ge (1 + \sqrt{y})^2</math>, with equality happening in the earlier inequality when <math>x = 1</math> and equality happening in the latter inequality when <math>y = 1</math>. Because <math>x,y > 0</math>, | By the [[Cauchy-Schwarz Inequality]], <math>(1+1)(1+x) \ge (1 + \sqrt{x})^2</math> and <math>(1+1)(1+y) \ge (1 + \sqrt{y})^2</math>, with equality happening in the earlier inequality when <math>x = 1</math> and equality happening in the latter inequality when <math>y = 1</math>. Because <math>x,y > 0</math>, | ||
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Therefore, since <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y})</math> and <math>\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}</math> and <math>\sqrt{\frac{1}{(x+1)(y+1)}} \ge \frac{2}{x+y+2}</math>, we must have <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac{2}{x+y+2}</math>, with equality happening when <math>x = y = 1</math>. | Therefore, since <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y})</math> and <math>\frac12 \cdot (\frac{1}{1+x} + \frac{1}{1+y}) \ge \sqrt{\frac{1}{(1+x)(1+y)}}</math> and <math>\sqrt{\frac{1}{(x+1)(y+1)}} \ge \frac{2}{x+y+2}</math>, we must have <math>\frac{1}{(1 + \sqrt{x})^2} + \frac{1}{(1 + \sqrt{y})^2} \ge \frac{2}{x+y+2}</math>, with equality happening when <math>x = y = 1</math>. | ||
− | Solution | + | ==Solution 2== |
Let <math>f(j)=\frac{1}{(1+\sqrt{j})^2}</math> | Let <math>f(j)=\frac{1}{(1+\sqrt{j})^2}</math> | ||
Since this function is concave up, according to Jensen's inequality, we can get <math>\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})</math> which means <math>f(x)+f(y)\geq 2f(\frac{x+y}{2})</math>. | Since this function is concave up, according to Jensen's inequality, we can get <math>\frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})</math> which means <math>f(x)+f(y)\geq 2f(\frac{x+y}{2})</math>. |
Latest revision as of 23:09, 3 December 2021
Contents
[hide]Problem
Prove that for every positive reals and
,
Solution 1
By the Cauchy-Schwarz Inequality, and
, with equality happening in the earlier inequality when
and equality happening in the latter inequality when
. Because
,
By the AM-GM Inequality, we know that
. For the equality case,
, so
. Additionally, by the AM-GM Inequality,
. For the equality case,
, so
. Because
,
Therefore, since
and
and
, we must have
, with equality happening when
.
Solution 2
Let
Since this function is concave up, according to Jensen's inequality, we can get
which means
.
In this problem, it turns into
.The conclusion we try to find is that
So we can see that
.
Take reciprocal for both sides we can get
.
Take RHS,
.
Now we have to prove that
.
which turns to
. It is always correct according to
inequality, it happens when
.
~bluesoul
See Also
2008 Indonesia MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 3 |
All Indonesia MO Problems and Solutions |