Determine the smallest possible value of the sum $S (a, b, c) = \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}$ where $a, b, c$ are three positive real numbers with $a^2 + b^2 + c^2 = 1$
Problem
Source: Germany Federal - Bundeswettbewerb Mathematik 2019 round 2, p2
Tags: inequalities, min, algebra
16.04.2020 13:48
6? By using AM-GM
16.04.2020 13:55
$$LHS=\sqrt{\sum\frac{a^2b^2}{c^2}+2\sum a^2}\geq\sqrt{\sum\frac{a^2b^2}{b^2}+2}=\sqrt{3}$$where we used rearrangement inequality
16.04.2020 14:52
parmenides51 wrote: Determine the smallest possible value of the sum $S (a, b, c) = \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}$ where $a, b, c$ are three positive real numbers with $a^2 + b^2 + c^2 = 1$ Stronger \[\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \geqslant \sqrt{4(a^2+b^2+c^2)-ab-bc-ca}.\]
16.04.2020 15:48
parmenides51 wrote: Determine the smallest possible value of the sum $S (a, b, c) = \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}$ where $a, b, c$ are three positive real numbers with $a^2 + b^2 + c^2 = 1$ Resemble to France MO 2005.
16.04.2020 16:15
Nguyenhuyen_AG wrote: parmenides51 wrote: Determine the smallest possible value of the sum $S (a, b, c) = \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}$ where $a, b, c$ are three positive real numbers with $a^2 + b^2 + c^2 = 1$ Stronger \[\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} \geqslant \sqrt{4(a^2+b^2+c^2)-ab-bc-ca}.\] Nice, but easy! Let $c=\min\{a,b,c\}$. It's equivalent to: $${c}^{4} \left( a-b \right) ^{2} \left\{ 3\,{c}^{2}+5\, \left( a-c \right) \left( -c+b \right) \right\} +2\, \left( -2\,c+b+a \right) ^{3}{c}^{5}+M(a-c)(b-c)\geqq 0$$But: $M=3\,{c}^{6}+35\,{c}^{4} \left( a-c \right) \left( -c+b \right) \\+20\, \left( a-c \right) \left( -c+b \right) \left( -2\,c+b+a \right) {c} ^{3}\\+ \left( a-c \right) \left( -c+b \right) \left( a-5\,c+4\,b \right) \left( 4\,a-5\,c+b \right) {c}^{2}\\+4\, \left( a-c \right) ^{ 2} \left( -c+b \right) ^{2} \left( -2\,c+b+a \right) c\\+ \left( a-c \right) ^{3} \left( -c+b \right) ^{3} \geqq 0$ Done!
06.05.2023 13:04
This problem is also there in evan otis application this year!!!
04.01.2025 05:30
also we can solve it with lagrange multipliers,if a,b,c are not equal the system of equations is impossible.and the minimum is sqrt(3) for a=b=c=1/sqrt(3)
04.01.2025 09:52
ektorasmiliotis wrote: also we can solve it with lagrange multipliers,if a,b,c are not equal the system of equations is impossible.and the minimum is sqrt(3) for a=b=c=1/sqrt(3) You need to prove the existence of the minimum.