Difference between revisions of "1998 IMO Shortlist Problems/A3"
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<math>\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\geq\frac{3}{4}</math> | <math>\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\geq\frac{3}{4}</math> | ||
| − | Solution1 | + | ==Solution1== |
Using Titu's Lemma, we rewrite the equation, getting that <math>\frac{(x^2+y^2+z^2)^2}{3xyz+2(xy+xz+yz)+x+y+z}\geq\frac{3}{4}</math>. | Using Titu's Lemma, we rewrite the equation, getting that <math>\frac{(x^2+y^2+z^2)^2}{3xyz+2(xy+xz+yz)+x+y+z}\geq\frac{3}{4}</math>. | ||
which means that <math>4(x^2+y^2+z^2)^2\geq 9+6(xy+yz+xz)+3(x+y+z)</math>. Applying AM-GM inequality, we can get that <math>x^2+y^2+z^2\geq3(x^2y^2z^2)^\frac{1}{3}=3</math> Which means the left side is greater than or equal to 36. | which means that <math>4(x^2+y^2+z^2)^2\geq 9+6(xy+yz+xz)+3(x+y+z)</math>. Applying AM-GM inequality, we can get that <math>x^2+y^2+z^2\geq3(x^2y^2z^2)^\frac{1}{3}=3</math> Which means the left side is greater than or equal to 36. | ||
Now let's observe the right side. Using C-S inequality we can easily get that <math>(x^2+y^2+z^2)^2\geq(xy+yz+xz)^2</math>. So <math>2(x^2+y^2+z^2)^2\geq 6(xy+yz+xz)</math> In the end, we can get that <math>(x^2+y^2+z^2)^2\geq(xy+yz+xz)^2=2xyz+x^2y^2+x^2z^2+y^2z^2\geq 2xyz(x+y+z)+3(x^4y^4z^4)^\frac{1}{3}=2xyz+3</math> Since <math>x^2+y^2+z^2\geq3</math>, we can get that <math>x+y+z\leq3</math> | Now let's observe the right side. Using C-S inequality we can easily get that <math>(x^2+y^2+z^2)^2\geq(xy+yz+xz)^2</math>. So <math>2(x^2+y^2+z^2)^2\geq 6(xy+yz+xz)</math> In the end, we can get that <math>(x^2+y^2+z^2)^2\geq(xy+yz+xz)^2=2xyz+x^2y^2+x^2z^2+y^2z^2\geq 2xyz(x+y+z)+3(x^4y^4z^4)^\frac{1}{3}=2xyz+3</math> Since <math>x^2+y^2+z^2\geq3</math>, we can get that <math>x+y+z\leq3</math> | ||
Now consider the right side <math>9+6(xy+yz+xz)+3(x+y+z)\leq9+6*3+3*3=36</math>, the left side <math>4(x^2+y^2+z^2)^2\geq 4*9=36</math> so left side is always larger or equal to right sides and we are done ~bluesoul | Now consider the right side <math>9+6(xy+yz+xz)+3(x+y+z)\leq9+6*3+3*3=36</math>, the left side <math>4(x^2+y^2+z^2)^2\geq 4*9=36</math> so left side is always larger or equal to right sides and we are done ~bluesoul | ||
Revision as of 02:12, 19 December 2021
Let 
be positive real numbers such that 
. Prove that
Solution1
Using Titu's Lemma, we rewrite the equation, getting that 
.
which means that 
. Applying AM-GM inequality, we can get that 
Which means the left side is greater than or equal to 36.
Now let's observe the right side. Using C-S inequality we can easily get that 
. So 
In the end, we can get that 
Since 
, we can get that 
Now consider the right side 
, the left side 
so left side is always larger or equal to right sides and we are done ~bluesoul
