# Difference between revisions of "1995 IMO Problems/Problem 2"

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=== Solution 4 === | === Solution 4 === | ||

− | After the setting <math>a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z},</math> and as <math>abc=1</math> so <math>\frac{1}{a} | + | After the setting <math>a=\frac{1}{x}, b=\frac{1}{y}, c=\frac{1}{z},</math> and as <math>abc=1</math> so <math>\left(\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1} {c}=1\right)</math> concluding <math>x y z=1 .</math> |

− | + | <math>\textsf{Claim}:</math> | |

− | \ | + | <cmath> |

− | + | \frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | |

− | + | </cmath> | |

By Titu Lemma, | By Titu Lemma, | ||

− | + | <cmath> | |

\implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)^{2}}{2(x+y+z)} | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)^{2}}{2(x+y+z)} | ||

− | + | </cmath> | |

− | + | <cmath> | |

\implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)}{2} | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{(x+y+z)}{2} | ||

− | + | </cmath> | |

− | Now by AM-GM we know that | + | Now by AM-GM we know that <cmath> (x+y+z)\geq3\sqrt[3]{xyz} |

− | + | </cmath>and <math>xyz=1</math> which concludes to <math>\implies (x+y+z)\geq3\sqrt[3]{1}</math> | |

Therefore we get | Therefore we get | ||

− | + | <cmath> | |

\implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | \implies\frac{x^{2}}{y+z}+\frac{y^{2}}{z+x}+\frac{z^{2}}{x+y} \geq \frac{3}{2} | ||

− | + | </cmath>Hence our claim is proved ~~ Aritra12 | |

=== Solution 5 === | === Solution 5 === |

## Latest revision as of 16:30, 19 June 2021

## Contents

## Problem

(*Nazar Agakhanov, Russia*)
Let be positive real numbers such that . Prove that

## Solution

### Solution 1

We make the substitution , , . Then Since and are similarly sorted sequences, it follows from the Rearrangement Inequality that By the Power Mean Inequality, Symmetric application of this argument yields Finally, AM-GM gives us as desired.

### Solution 2

We make the same substitution as in the first solution. We note that in general, It follows that and are similarly sorted sequences. Then by Chebyshev's Inequality, By AM-GM, , and by Nesbitt's Inequality, The desired conclusion follows.

### Solution 3

Without clever substitutions: By Cauchy-Schwarz, Dividing by gives by AM-GM.

### Solution 3b

Without clever notation: By Cauchy-Schwarz,

Dividing by and noting that by AM-GM gives as desired.

### Solution 4

After the setting and as so concluding

By Titu Lemma, Now by AM-GM we know that and which concludes to

Therefore we get

Hence our claim is proved ~~ Aritra12

### Solution 5

Proceed as in Solution 1, to arrive at the equivalent inequality But we know that by AM-GM. Furthermore, by Cauchy-Schwarz, and so dividing by gives as desired.

### Solution 6

Without clever substitutions, and only AM-GM!

Note that . The cyclic sum becomes . Note that by AM-GM, the cyclic sum is greater than or equal to . We now see that we have the three so we must be on the right path. We now only need to show that . Notice that by AM-GM, , , and . Thus, we see that , concluding that

### Solution 7 from Brilliant Wiki (Muirheads) =

https://brilliant.org/wiki/muirhead-inequality/

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*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*