Difference between revisions of "1998 IMO Shortlist Problems/A3"
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Let <math>x,y,z</math> be positive real numbers such that <math>xyz=1</math>. Prove that | Let <math>x,y,z</math> be positive real numbers such that <math>xyz=1</math>. Prove that | ||
<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== | ||
+ | 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. | ||
+ | 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+y+z)+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 | ||
+ | |||
+ | ==Solution2== | ||
+ | WLOG, we assume that <math>x\leq y\leq z</math>, so <math>\frac{1}{(1+y)(1+z)}\leq \frac{1}{(1+x)(1+z)}\leq \frac{1}{(1+x)(1+y)}</math>. Now we can apply Chebyshev's inequality, getting that <math>\frac{x^3}{(1+y)(1+z)}+\frac{y^3}{(1+x)(1+z)}+\frac{z^3}{(1+x)(1+y)}\geq \frac{(x^3+y^3+z^3)(3+x+y+z)}{3(xy+xz+yz+x+y+z+1+xyz)}</math>. Now we can observe that all three terms, <math>(x^3+y^3+z^3);(x+y+z);(xy+yz+xz)</math> get their minimum value <math>3</math> while <math>x=y=z</math>. Which means the minimum value of the original expression holds while <math>x=y=z=1</math> getting <math>\frac{3}{4}</math> and we are done ~bluesoul |
Latest revision as of 11:44, 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
Solution2
WLOG, we assume that , so . Now we can apply Chebyshev's inequality, getting that . Now we can observe that all three terms, get their minimum value while . Which means the minimum value of the original expression holds while getting and we are done ~bluesoul