Question

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

Answer 1

@mukushla @KingGeorge

Answer 2

\[\mathsf{\text{Prove}\quad\frac{a}{b+c+1} + \frac{b}{a+c+1} + \frac{c}{a+b+1} + \left(1-a\right)\left(1-b\right)\left(1-c\right)\le1}\]


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 (1-a), (1-b),(1-c) will lie between (0,1)

and so

(1-a)(1-b)(1-c) 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

Answer 3

(1-a)(1-b)(1-c) will be approx. = 0


this is because product of any three decimals <1 and > 0 their product will be very close to 0

Answer 4

@Ishaan94

Can this help u a little ............... ?

Answer 5

another solution using calculus

Answer 6

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.

Answer 7

@mukushla
It is really a very nice solution