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\[\mathsf{a,b,c \in \left[0,1\right]}\]\[\mathsf{\text{Prove}\quad\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} + \left(1a\right)\left(1b\right)\left(1c\right)\le1}.\]
 one year ago
 one year ago
\[\mathsf{a,b,c \in \left[0,1\right]}\]\[\mathsf{\text{Prove}\quad\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} + \left(1a\right)\left(1b\right)\left(1c\right)\le1}.\]
 one year ago
 one year ago

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Ishaan94Best ResponseYou've already chosen the best response.0
@mukushla @KingGeorge
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.1
\[\mathsf{\text{Prove}\quad\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} + \left(1a\right)\left(1b\right)\left(1c\right)\le1}\] Because \[\mathsf{a,b,c \in \left[0,1\right]}\] It can be easily seen that for a,b,c = 0 or 1 the LHS becomes = 1 Now in all other remaining cases i.e. a,b,c belong to (0,1) each of the fraction will lie in (0,1/3) and each of (1a), (1b),(1c) will lie between (0,1) and so (1a)(1b)(1c) will be approx. = 0 and so the whole LHS < 1 I know that it is a very bad proof But it is just a try. Sorry for this bad attempt
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.1
(1a)(1b)(1c) will be approx. = 0 this is because product of any three decimals <1 and > 0 their product will be very close to 0
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.1
@Ishaan94 Can this help u a little ............... ?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
another solution using calculus
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.0
thank you mukushla. i do like the solution, thanks a lot. and thank you vishwesh i had already thought of the way the you did but wasn't able to generate a concrete solution.
 one year ago

vishweshshrimali5Best ResponseYou've already chosen the best response.1
@mukushla It is really a very nice solution
 one year ago
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