# Difference between revisions of "2020 IMO Problems/Problem 2"

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<cmath>\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2</cmath> | <cmath>\implies a^a b^b c^c d^d \le a^2 +b^2 +c^2 +d^2</cmath> | ||

− | So, <cmath>(a+2b+3c+4d) a^ab^bc^ | + | So, <cmath>(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) </cmath> |

Now notice that | Now notice that | ||

− | <cmath>a+2b+3c+4d \text{ will be less then the following | + | <cmath>a+2b+3c+4d \text{ will be less then the following expressions (and the reason is written to the right)} </cmath> |

<cmath>a+2b+3c+3d,\text{as } d\le b</cmath> | <cmath>a+2b+3c+3d,\text{as } d\le b</cmath> | ||

<cmath>3a+3b+3c+d, \text{as } d\le a</cmath> | <cmath>3a+3b+3c+d, \text{as } d\le a</cmath> | ||

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So, we get | So, we get | ||

− | <cmath>(a+2b+3c+4d)(a^2+b^2+c^2+d^2) | + | <cmath>\begin{split}&~~~~(a+2b+3c+4d)(a^2+b^2+c^2+d^2) \\ |

− | + | &= a^2(a+2b+3c+4d)+b^2(a+2b+3c+4d)+c^2 (a+2b+3c+4d) +d^2 (a+2b+3c+4d)\\ | |

− | + | &\le a^2(a+3b+3c+3d)+b^2(3a+b+3c+3d)+c^2 (3a+3b+c+3d) +d^2 (3a+3b+3c+d)\\ | |

− | + | &<(a+b+c+d)^3 \\&=1\end{split}</cmath> | |

− | |||

− | |||

Now, for equality we must have <math>a=b=c=d=\frac{1}{4}</math> | Now, for equality we must have <math>a=b=c=d=\frac{1}{4}</math> | ||

− | In that case we get <cmath>(a+2b+3c+4d) a^ab^bc^ | + | In that case we get <cmath>(a+2b+3c+4d) a^ab^bc^cd^d \le (a+2b+3c+4d)(a^2+b^2+c^2+d^2) =\frac{5}{8} <1</cmath> |

~ftheftics | ~ftheftics |

## Revision as of 09:36, 2 October 2020

Problem 2. The real numbers are such that and . Prove that

## Solution

Using Weighted AM-GM we get

So,

Now notice that

So, we get

Now, for equality we must have

In that case we get

~ftheftics